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# Learn UNIT IV: Geometry with Free Lessons & Tips

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To construct a triangle read more

To construct a triangle $\fn_cm&space;$$&space;PQR&space;$$&space;similar&space;to&space;triangle&space;$$&space;ABC&space;$$,&space;with&space;$$&space;PQ&space;=&space;8&space;$$&space;cm,&space;we&space;follow&space;these&space;steps:&space;1.&space;Draw&space;a&space;line&space;segment&space;$$&space;PQ&space;$$&space;of&space;length&space;$$&space;8&space;$$&space;cm.&space;2.&space;At&space;point&space;$$&space;P&space;$$,&space;construct&space;an&space;angle&space;equal&space;to&space;angle&space;$$&space;BAC&space;$$&space;using&space;a&space;protractor.&space;3.&space;From&space;point&space;$$&space;P&space;$$,&space;draw&space;a&space;line&space;segment&space;$$&space;PR&space;$$&space;making&space;this&space;angle.&space;4.&space;Extend&space;$$&space;PQ&space;$$&space;beyond&space;$$&space;P&space;$$&space;to&space;point&space;$$&space;R&space;$$&space;such&space;that&space;$$&space;PR&space;=&space;8&space;$$&space;cm.&space;5.&space;Through&space;points&space;$$&space;Q&space;$$&space;and&space;$$&space;R&space;$$,&space;draw&space;lines&space;parallel&space;to&space;$$&space;BC&space;$$.&space;6.&space;Label&space;the&space;intersection&space;of&space;these&space;parallel&space;lines&space;with&space;$$&space;PQ&space;$$&space;as&space;point&space;$$&space;S&space;$$&space;and&space;with&space;$$&space;PR&space;$$&space;as&space;point&space;$$&space;T&space;$$.&space;7.&space;Join&space;points&space;$$&space;S&space;$$&space;and&space;$$&space;T&space;$$&space;to&space;$$&space;R&space;$$.&space;8.&space;$$&space;\triangle&space;PQR&space;$$&space;is&space;the&space;required&space;triangle.$

$\fn_cm&space;Justification:&space;-&space;Angle&space;$$&space;P&space;$$&space;is&space;constructed&space;equal&space;to&space;angle&space;$$&space;A&space;$$&space;because&space;corresponding&space;angles&space;in&space;similar&space;triangles&space;are&space;equal.&space;-&space;$$&space;PQ&space;$$&space;is&space;chosen&space;as&space;$$&space;8&space;$$&space;cm&space;to&space;make&space;the&space;scale&space;factor&space;between&space;$$&space;PQ&space;$$&space;and&space;$$&space;AB&space;$$&space;equal&space;to&space;$$&space;\frac{8}{6}&space;=&space;\frac{4}{3}&space;$$,&space;ensuring&space;similarity&space;between&space;$$&space;\triangle&space;ABC&space;$$&space;and&space;$$&space;\triangle&space;PQR&space;$$.&space;-&space;The&space;lines&space;parallel&space;to&space;$$&space;BC&space;$$&space;through&space;points&space;$$&space;Q&space;$$&space;and&space;$$&space;R&space;$$&space;ensure&space;that&space;the&space;corresponding&space;sides&space;of&space;$$&space;\triangle&space;PQR&space;$$&space;are&space;parallel&space;to&space;the&space;corresponding&space;sides&space;of&space;$$&space;\triangle&space;ABC&space;$$,&space;which&space;is&space;a&space;property&space;of&space;similar&space;triangles.$

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To construct the tangents from point read more

To construct the tangents from point $\fn_cm&space;\large&space;$$&space;A&space;$$&space;to&space;the&space;circle&space;passing&space;through&space;points&space;$$&space;B&space;$$,&space;$$&space;C&space;$$,&space;and&space;$$&space;D&space;$$,&space;we&space;follow&space;these&space;steps:&space;1.&space;Construct&space;triangle&space;$$&space;ABC&space;$$&space;with&space;$$&space;AB&space;=&space;6&space;$$&space;cm,&space;$$&space;BC&space;=&space;8&space;$$&space;cm,&space;and&space;$$&space;\angle&space;B&space;=&space;90^\circ&space;$$.&space;2.&space;Draw&space;the&space;altitude&space;$$&space;BD&space;$$&space;from&space;vertex&space;$$&space;B&space;$$&space;to&space;the&space;hypotenuse&space;$$&space;AC&space;$$.&space;3.&space;Locate&space;the&space;midpoint&space;$$&space;M&space;$$&space;of&space;$$&space;BD&space;$$.&space;4.&space;Draw&space;a&space;circle&space;passing&space;through&space;points&space;$$&space;B&space;$$,&space;$$&space;C&space;$$,&space;and&space;$$&space;D&space;$$.&space;This&space;circle&space;has&space;$$&space;BD&space;$$&space;as&space;a&space;diameter.&space;5.&space;Draw&space;the&space;perpendicular&space;bisector&space;of&space;$$&space;BD&space;$$,&space;which&space;passes&space;through&space;point&space;$$&space;M&space;$$&space;and&space;the&space;center&space;of&space;the&space;circle&space;(let's&space;call&space;it&space;$$&space;O&space;$$).&space;6.&space;Construct&space;the&space;perpendicular&space;from&space;point&space;$$&space;A&space;$$&space;to&space;line&space;$$&space;MO&space;$$,&space;intersecting&space;it&space;at&space;point&space;$$&space;E&space;$$.&space;7.&space;$$&space;EA&space;$$&space;and&space;$$&space;EA'&space;$$&space;are&space;the&space;required&space;tangents&space;from&space;point&space;$$&space;A&space;$$&space;to&space;the&space;circle,&space;where&space;$$&space;A'&space;$$&space;is&space;the&space;reflection&space;of&space;$$&space;A&space;$$&space;across&space;line&space;$$&space;MO&space;$$.$

$\fn_cm&space;\large&space;Justification:&space;-&space;The&space;circle&space;passing&space;through&space;$$&space;B&space;$$,&space;$$&space;C&space;$$,&space;and&space;$$&space;D&space;$$&space;has&space;$$&space;BD&space;$$&space;as&space;a&space;diameter,&space;which&space;implies&space;that&space;$$&space;\angle&space;CBD&space;=&space;90^\circ&space;$$&space;and&space;$$&space;\angle&space;CDB&space;=&space;90^\circ&space;$$.&space;-&space;The&space;tangents&space;from&space;an&space;external&space;point&space;to&space;a&space;circle&space;are&space;equal&space;in&space;length.&space;So,&space;$$&space;EA&space;=&space;EA'&space;$$.&space;-&space;The&space;tangent&space;$$&space;EA&space;$$&space;is&space;constructed&space;perpendicular&space;to&space;the&space;radius&space;of&space;the&space;circle&space;at&space;the&space;point&space;of&space;tangency.&space;Similarly,&space;$$&space;EA'&space;$$&space;is&space;perpendicular&space;to&space;the&space;radius&space;at&space;the&space;point&space;of&space;tangency.&space;-&space;By&space;construction,&space;$$&space;EA&space;$$&space;and&space;$$&space;EA'&space;$$&space;are&space;tangents&space;from&space;point&space;$$&space;A&space;$$&space;to&space;the&space;circle&space;passing&space;through&space;points&space;$$&space;B&space;$$,&space;$$&space;C&space;$$,&space;and&space;$$&space;D&space;$$.$

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To prove that the above equation read more

To prove that  the above equation $\fn_cm&space;$$&space;AD&space;=&space;BC&space;$$&space;in&space;trapezium&space;$$&space;ABCD&space;$$,&space;where&space;$$&space;AB&space;\parallel&space;DC&space;$$&space;and&space;$$&space;AAED&space;\sim&space;ABEC&space;$$,&space;we&space;can&space;use&space;the&space;properties&space;of&space;similar&space;triangles.&space;Given:&space;Trapezium&space;$$&space;ABCD&space;$$&space;with&space;$$&space;AB&space;\parallel&space;DC&space;$$,&space;and&space;$$&space;AAED&space;\sim&space;ABEC&space;$$.&space;First,&space;let's&space;consider&space;triangles&space;$$&space;AED&space;$$&space;and&space;$$&space;BEC&space;$$.&space;Since&space;$$&space;AAED&space;\sim&space;ABEC&space;$$,&space;we&space;have:&space;$&space;\frac{AD}{AB}&space;=&space;\frac{AE}{BE}&space;=&space;\frac{ED}{EC}&space;$&space;---(1)&space;Now,&space;let's&space;consider&space;triangles&space;$$&space;ADE&space;$$&space;and&space;$$&space;BCE&space;$$.&space;Since&space;$$&space;AB&space;\parallel&space;DC&space;$$,&space;we&space;have:&space;$&space;\angle&space;ADE&space;=&space;\angle&space;BCE&space;$&space;(corresponding&space;angles)&space;$&space;\angle&space;AED&space;=&space;\angle&space;BEC&space;$&space;(alternate&space;angles)&space;Therefore,&space;by&space;Angle-Angle&space;similarity&space;criterion,&space;triangles&space;$$&space;ADE&space;$$&space;and&space;$$&space;BCE&space;$$&space;are&space;similar.&space;From&space;the&space;similarity&space;of&space;triangles&space;$$&space;ADE&space;$$&space;and&space;$$&space;BCE&space;$$,&space;we&space;have:&space;$&space;\frac{AD}{BC}&space;=&space;\frac{AE}{BE}&space;=&space;\frac{ED}{EC}&space;$&space;---(2)$

$\fn_cm&space;Comparing&space;equations&space;(1)&space;and&space;(2),&space;we&space;see&space;that:&space;$&space;\frac{AD}{AB}&space;=&space;\frac{AD}{BC}&space;$&space;Since&space;$$&space;AB&space; eq&space;0&space;$$,&space;we&space;can&space;conclude&space;that&space;$$&space;AD&space;=&space;BC&space;$$.&space;Thus,&space;we&space;have&space;proved&space;that&space;in&space;trapezium&space;$$&space;ABCD&space;$$,&space;where&space;$$&space;AB&space;\parallel&space;DC&space;$$&space;and&space;$$&space;AAED&space;\sim&space;ABEC&space;$$,&space;$$&space;AD&space;=&space;BC&space;$$.$

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Let's denote the area of triangle read more

Let's denote the area of triangle $$$&space;ABC&space;$$&space;as&space;$$&space;[ABC]&space;$$.&space;Given&space;that&space;$$&space;PQ&space;\parallel&space;BC&space;$$&space;divides&space;$$&space;\triangle&space;ABC&space;$$&space;into&space;two&space;parts&space;of&space;equal&space;area.&space;Let's&space;denote&space;the&space;areas&space;of&space;$$&space;\triangle&space;APQ&space;$$&space;and&space;$$&space;\triangle&space;AQP&space;$$&space;as&space;$$&space;[APQ]&space;$$&space;and&space;$$&space;[AQP]&space;$$&space;respectively.&space;Since&space;$$&space;[APQ]&space;=&space;[AQP]&space;$$,&space;we&space;have:&space;$&space;[ABC]&space;=&space;[APQ]&space;+&space;[AQP]&space;$&space;$&space;[ABC]&space;=&space;2&space;\cdot&space;[APQ]&space;$&space;Since&space;$$&space;[APQ]&space;=&space;[AQP]&space;$$,&space;we&space;can&space;say:&space;$&space;[APQ]&space;=&space;\frac{1}{2}&space;[ABC]&space;$$

$\fn_cm&space;Now,&space;let's&space;express&space;$$&space;[ABC]&space;$$&space;in&space;terms&space;of&space;the&space;areas&space;of&space;triangles&space;$$&space;APQ&space;$$,&space;$$&space;PBQ&space;$$,&space;and&space;$$&space;AQB&space;$$:&space;$&space;[ABC]&space;=&space;[APQ]&space;+&space;[PBQ]&space;+&space;[AQB]&space;$&space;Given&space;that&space;$$&space;[APQ]&space;=&space;[AQB]&space;=&space;\frac{1}{2}&space;[ABC]&space;$$,&space;we&space;can&space;substitute&space;these&space;values:&space;$&space;[ABC]&space;=&space;\frac{1}{2}&space;[ABC]&space;+&space;[PBQ]&space;+&space;\frac{1}{2}&space;[ABC]&space;$&space;$&space;[ABC]&space;=&space;[PBQ]&space;+&space;[ABC]&space;$&space;$&space;[PBQ]&space;=&space;0&space;$&space;This&space;implies&space;that&space;triangle&space;$$&space;PBQ&space;$$&space;has&space;zero&space;area.&space;Since&space;the&space;area&space;of&space;a&space;triangle&space;cannot&space;be&space;negative,&space;this&space;means&space;that&space;the&space;points&space;$$&space;P&space;$$,&space;$$&space;B&space;$$,&space;and&space;$$&space;Q&space;$$&space;are&space;collinear.&space;Therefore,&space;$$&space;BP&space;$$&space;is&space;a&space;line&space;segment&space;parallel&space;to&space;$$&space;AC&space;$$&space;and&space;$$&space;AB&space;$$,&space;hence:&space;$&space;\frac{BP}{AB}&space;=&space;\frac{BP}{BP&space;+&space;PQ}&space;=&space;\frac{BP}{BP&space;+&space;BC}&space;=&space;\frac{1}{2}&space;$&space;So,&space;$$&space;\frac{BP}{AB}&space;=&space;\frac{1}{2}&space;$$.$

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Given: △ABC is an isosceles triangle with AB=AC, and D is any point on BC We'll use the Law of Cosines to prove the given relation. read more

Given: △ABC is an isosceles triangle with AB=AC, and D is any point on BC

We'll use the Law of Cosines to prove the given relation.
$\fn_cm&space;In&space;$$&space;\triangle&space;ABD&space;$$:&space;$&space;(AB)^2&space;=&space;(AD)^2&space;+&space;(BD)^2&space;-&space;2&space;\cdot&space;AD&space;\cdot&space;BD&space;\cdot&space;\cos(\angle&space;ABD)&space;$&space;In&space;$$&space;\triangle&space;ACD&space;$$:&space;$&space;(AC)^2&space;=&space;(AD)^2&space;+&space;(CD)^2&space;-&space;2&space;\cdot&space;AD&space;\cdot&space;CD&space;\cdot&space;\cos(\angle&space;ACD)&space;$&space;Since&space;$$&space;\triangle&space;ABC&space;$$&space;is&space;isosceles,&space;$$&space;\angle&space;ABD&space;=&space;\angle&space;ACD&space;$$.&space;Given&space;that&space;$$&space;AB&space;=&space;AC&space;$$,&space;$$&space;(AB)^2&space;=&space;(AC)^2&space;$$.&space;Substituting&space;$$&space;(AB)^2&space;$$&space;and&space;$$&space;(AC)^2&space;$$&space;in&space;the&space;above&space;equations,&space;we&space;have:&space;$&space;(AD)^2&space;+&space;(BD)^2&space;-&space;2&space;\cdot&space;AD&space;\cdot&space;BD&space;\cdot&space;\cos(\angle&space;ABD)&space;=&space;(AD)^2&space;+&space;(CD)^2&space;-&space;2&space;\cdot&space;AD&space;\cdot&space;CD&space;\cdot&space;\cos(\angle&space;ABD)&space;$&space;Since&space;$$&space;\angle&space;ABD&space;=&space;\angle&space;ACD&space;$$,&space;$$&space;\cos(\angle&space;ABD)&space;=&space;\cos(\angle&space;ACD)&space;$$.$$\fn_cm&space;This&space;implies:&space;$&space;(BD)^2&space;-&space;2&space;\cdot&space;AD&space;\cdot&space;BD&space;\cdot&space;\cos(\angle&space;ABD)&space;=&space;(CD)^2&space;-&space;2&space;\cdot&space;AD&space;\cdot&space;CD&space;\cdot&space;\cos(\angle&space;ABD)&space;$&space;$&space;(BD)^2&space;-&space;2&space;\cdot&space;AD&space;\cdot&space;BD&space;\cdot&space;\cos(\angle&space;ABD)&space;-&space;(CD)^2&space;+&space;2&space;\cdot&space;AD&space;\cdot&space;CD&space;\cdot&space;\cos(\angle&space;ABD)&space;=&space;0&space;$&space;$&space;(BD)^2&space;-&space;(CD)^2&space;-&space;2&space;\cdot&space;AD&space;\cdot&space;BD&space;\cdot&space;\cos(\angle&space;ABD)&space;+&space;2&space;\cdot&space;AD&space;\cdot&space;CD&space;\cdot&space;\cos(\angle&space;ABD)&space;=&space;0&space;$&space;$&space;(BD)^2&space;-&space;(CD)^2&space;-&space;2&space;\cdot&space;AD&space;\cdot&space;(\cos(\angle&space;ABD)&space;\cdot&space;BD&space;-&space;\cos(\angle&space;ABD)&space;\cdot&space;CD)&space;=&space;0&space;$&space;$&space;(BD)^2&space;-&space;(CD)^2&space;-&space;2&space;\cdot&space;AD&space;\cdot&space;BD&space;-&space;CD&space;=&space;0&space;$&space;$&space;(BD)^2&space;-&space;(CD)^2&space;=&space;2&space;\cdot&space;AD&space;\cdot&space;BD&space;+&space;CD&space;\cdot&space;AD&space;$&space;$&space;(BD)^2&space;-&space;(CD)^2&space;=&space;(BD&space;+&space;CD)&space;\cdot&space;AD&space;$&space;$&space;(BD)^2&space;-&space;(AD)^2&space;=&space;BD&space;\cdot&space;CD&space;$&space;So,&space;$$&space;(AB)^2&space;-&space;(AD)^2&space;=&space;BD&space;\cdot&space;CD&space;$$,&space;as&space;required.$

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Given :: solution:: for above Question read more

Given :: solution:: for above Question $\fn_cm&space;In&space;trapezium&space;$$&space;ABCD&space;$$&space;with&space;$$&space;AB&space;\parallel&space;DC&space;$$,&space;let&space;$$&space;AC&space;$$&space;and&space;$$&space;BD&space;$$&space;intersect&space;at&space;point&space;$$&space;O&space;$$.&space;We&space;need&space;to&space;show&space;that&space;$$&space;\frac{OA}{OC}&space;=&space;\frac{OB}{OD}&space;$$&space;using&space;a&space;similarity&space;criterion&space;for&space;two&space;triangles.&space;Let's&space;consider&space;$$&space;\triangle&space;AOB&space;$$&space;and&space;$$&space;\triangle&space;DOC&space;$$:&space;1.&space;$$&space;\angle&space;AOB&space;$$&space;and&space;$$&space;\angle&space;DOC&space;$$&space;are&space;vertical&space;angles,&space;so&space;they&space;are&space;equal.&space;2.&space;$$&space;\angle&space;OAB&space;$$&space;and&space;$$&space;\angle&space;OCD&space;$$&space;are&space;alternate&space;interior&space;angles,&space;as&space;$$&space;AB&space;\parallel&space;DC&space;$$,&space;so&space;they&space;are&space;also&space;equal.&space;Since&space;$$&space;\triangle&space;AOB&space;$$&space;and&space;$$&space;\triangle&space;DOC&space;$$&space;have&space;two&space;angles&space;equal,&space;they&space;are&space;similar&space;by&space;the&space;Angle-Angle&space;(AA)&space;criterion.&space;In&space;similar&space;triangles,&space;the&space;ratios&space;of&space;corresponding&space;sides&space;are&space;equal.&space;Therefore:&space;$&space;\frac{OA}{OC}&space;=&space;\frac{OB}{OD}&space;$&space;So,&space;$$&space;\frac{OA}{OC}&space;=&space;\frac{OB}{OD}&space;$$,&space;as&space;required.$

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This is known as the Pythagorean Theorem, a fundamental principle in geometry. Let's denote the sides of the triangle as aa, b, and c, where c is the side opposite the right angle. read more

This is known as the Pythagorean Theorem, a fundamental principle in geometry. Let's denote the sides of the triangle as aa, b, and c, where c is the side opposite the right angle.
$\fn_cm&space;According&space;to&space;the&space;Pythagorean&space;Theorem,&space;in&space;a&space;right&space;triangle,&space;the&space;square&space;of&space;the&space;length&space;of&space;the&space;hypotenuse&space;($$&space;c&space;$$)&space;is&space;equal&space;to&space;the&space;sum&space;of&space;the&space;squares&space;of&space;the&space;lengths&space;of&space;the&space;other&space;two&space;sides&space;($$&space;a&space;$$&space;and&space;$$&space;b&space;$$).&space;Mathematically,&space;this&space;can&space;be&space;expressed&space;as:&space;$&space;c^2&space;=&space;a^2&space;+&space;b^2&space;$&space;Now,&space;let's&space;prove&space;the&space;converse:&space;if&space;in&space;a&space;triangle,&space;the&space;square&space;on&space;one&space;side&space;is&space;equal&space;to&space;the&space;sum&space;of&space;the&space;squares&space;on&space;the&space;other&space;two&space;sides,&space;then&space;the&space;angle&space;opposite&space;the&space;first&space;side&space;is&space;a&space;right&space;angle.$

$\fn_cm&space;Given:&space;$&space;c^2&space;=&space;a^2&space;+&space;b^2&space;$&space;We'll&space;use&space;the&space;Law&space;of&space;Cosines&space;to&space;prove&space;this.&space;In&space;a&space;triangle,&space;the&space;Law&space;of&space;Cosines&space;states:&space;$&space;c^2&space;=&space;a^2&space;+&space;b^2&space;-&space;2ab&space;\cos(C)&space;$&space;Where&space;$$&space;C&space;$$&space;is&space;the&space;angle&space;opposite&space;side&space;$$&space;c&space;$$.&space;Substituting&space;$$&space;c^2&space;=&space;a^2&space;+&space;b^2&space;$$&space;into&space;the&space;Law&space;of&space;Cosines,&space;we&space;have:&space;$&space;a^2&space;+&space;b^2&space;=&space;a^2&space;+&space;b^2&space;-&space;2ab&space;\cos(C)&space;$&space;$&space;0&space;=&space;-&space;2ab&space;\cos(C)&space;$$

$\fn_cm&space;Since&space;$$&space;a&space;$$&space;and&space;$$&space;b&space;$$&space;are&space;positive&space;lengths,&space;and&space;$$&space;\cos(C)&space;$$&space;is&space;a&space;trigonometric&space;function&space;with&space;values&space;between&space;-1&space;and&space;1,&space;the&space;only&space;way&space;for&space;the&space;equation&space;to&space;hold&space;true&space;is&space;if&space;$$&space;\cos(C)&space;=&space;0&space;$$.&space;For&space;$$&space;\cos(C)&space;=&space;0&space;$$,&space;$$&space;C&space;=&space;90^\circ&space;$$,&space;meaning&space;the&space;angle&space;opposite&space;side&space;$$&space;c&space;$$&space;is&space;a&space;right&space;angle.&space;So,&space;if&space;in&space;a&space;triangle,&space;the&space;square&space;on&space;one&space;side&space;is&space;equal&space;to&space;the&space;sum&space;of&space;the&space;squares&space;on&space;the&space;other&space;two&space;sides,&space;then&space;the&space;angle&space;opposite&space;the&space;first&space;side&space;is&space;a&space;right&space;angle.$

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To prove that the parallelogram circumscribing a circle is a rhombus, we first need to establish some properties: In a parallelogram, opposite sides are equal in length. In a parallelogram, opposite angles are equal in measure. Now, let's consider a parallelogram ABCD circumscribing a circle with center... read more

To prove that the parallelogram circumscribing a circle is a rhombus, we first need to establish some properties:

1. In a parallelogram, opposite sides are equal in length.
2. In a parallelogram, opposite angles are equal in measure.

Now, let's consider a parallelogram ABCD circumscribing a circle with center O. Let P, Q, R, and S be the points of tangency between the circle and the sides of the parallelogram, as shown below:
P-----Q
/       \
/         \
A-----------B
|     O     |
|           |
D-----------C
\         /
\       /
R-----S

$\fn_cm&space;We&space;need&space;to&space;prove&space;that&space;$$&space;ABCD&space;$$&space;is&space;a&space;rhombus,&space;meaning&space;that&space;all&space;four&space;sides&space;are&space;equal&space;in&space;length.&space;Since&space;$$&space;AB&space;\parallel&space;CD&space;$$&space;and&space;$$&space;AD&space;\parallel&space;BC&space;$$,&space;$$&space;ABCD&space;$$&space;is&space;a&space;parallelogram.&space;Now,&space;consider&space;triangle&space;$$&space;AOP&space;$$.&space;$$&space;OA&space;=&space;OP&space;$$&space;because&space;both&space;are&space;radii&space;of&space;the&space;same&space;circle.&space;Similarly,&space;$$&space;OB&space;=&space;OQ&space;$$,&space;$$&space;OC&space;=&space;OR&space;$$,&space;and&space;$$&space;OD&space;=&space;OS&space;$$.&space;Since&space;$$&space;AB&space;\parallel&space;CD&space;$$,&space;$$&space;AD&space;\parallel&space;BC&space;$$,&space;and&space;opposite&space;sides&space;of&space;a&space;parallelogram&space;are&space;equal,&space;we&space;have:&space;-&space;$$&space;OA&space;=&space;OC&space;$$&space;and&space;$$&space;OB&space;=&space;OD&space;$$&space;-&space;$$&space;OP&space;=&space;OR&space;$$&space;and&space;$$&space;OQ&space;=&space;OS&space;$$$

$\fn_cm&space;In&space;triangle&space;$$&space;AOP&space;$$&space;and&space;triangle&space;$$&space;COD&space;$$:&space;-&space;$$&space;OA&space;=&space;OC&space;$$&space;(radii&space;of&space;the&space;same&space;circle)&space;-&space;$$&space;OP&space;=&space;OC&space;$$&space;(tangents&space;drawn&space;from&space;an&space;external&space;point&space;are&space;equal&space;in&space;length)&space;-&space;$$&space;\angle&space;AOP&space;=&space;\angle&space;COD&space;$$&space;(vertically&space;opposite&space;angles)&space;By&space;the&space;Side-Angle-Side&space;(SAS)&space;congruence&space;criterion,&space;$$&space;\triangle&space;AOP&space;$$&space;is&space;congruent&space;to&space;$$&space;\triangle&space;COD&space;$$.&space;Similarly,&space;by&space;comparing&space;other&space;pairs&space;of&space;triangles,&space;we&space;can&space;show&space;that&space;$$&space;\triangle&space;BOP&space;$$&space;is&space;congruent&space;to&space;$$&space;\triangle&space;DOS&space;$$,&space;$$&space;\triangle&space;COQ&space;$$&space;is&space;congruent&space;to&space;$$&space;\triangle&space;ROS&space;$$,&space;and&space;$$&space;\triangle&space;DOQ&space;$$&space;is&space;congruent&space;to&space;$$&space;\triangle&space;BOS&space;$$.$

$\fn_cm&space;Since&space;congruent&space;triangles&space;have&space;equal&space;corresponding&space;sides,&space;we&space;conclude&space;that:&space;-&space;$$&space;AB&space;=&space;CD&space;$$&space;-&space;$$&space;AD&space;=&space;BC&space;$$&space;-&space;$$&space;AB&space;=&space;AD&space;$$&space;and&space;$$&space;BC&space;=&space;CD&space;$$&space;Therefore,&space;all&space;four&space;sides&space;of&space;the&space;parallelogram&space;$$&space;ABCD&space;$$&space;are&space;equal&space;in&space;length,&space;making&space;it&space;a&space;rhombus.$

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Let's denote the midpoint of BC as M. To prove that the tangent to the circle at P bisects BC we need to show that BM=MC. read more

Let's denote the midpoint of BC as M. To prove that the tangent to the circle at P bisects BC we need to show that BM=MC.$Given:&space;-&space;$$&space;\angle&space;B&space;=&space;90^\circ&space;$$&space;-&space;$$&space;AB&space;$$&space;is&space;the&space;diameter&space;of&space;the&space;circle&space;with&space;center&space;$$&space;O&space;$$&space;-&space;The&space;circle&space;intersects&space;the&space;hypotenuse&space;$$&space;AC&space;$$&space;at&space;point&space;$$&space;P&space;$$&space;Since&space;$$&space;AB&space;$$&space;is&space;the&space;diameter&space;of&space;the&space;circle,&space;$$&space;\angle&space;APB&space;=&space;90^\circ&space;$$&space;(angle&space;subtended&space;by&space;a&space;diameter&space;is&space;90&space;degrees).&space;Since&space;$$&space;\angle&space;B&space;=&space;90^\circ&space;$$,&space;$$&space;\angle&space;APB&space;=&space;90^\circ&space;$$,&space;and&space;$$&space;\angle&space;ABC&space;=&space;90^\circ&space;$$,&space;$$&space;\triangle&space;ABP&space;$$&space;is&space;a&space;right&space;triangle.$$\fn_cm&space;According&space;to&space;Thales'&space;theorem,&space;if&space;$$&space;AC&space;$$&space;is&space;a&space;diameter&space;of&space;the&space;circle,&space;and&space;$$&space;P&space;$$&space;lies&space;on&space;the&space;circle,&space;then&space;$$&space;\angle&space;APC&space;=&space;90^\circ&space;$$.&space;So,&space;$$&space;\angle&space;APC&space;=&space;\angle&space;ABC&space;=&space;90^\circ&space;$$.&space;Since&space;$$&space;\angle&space;APC&space;=&space;90^\circ&space;$$&space;and&space;$$&space;\angle&space;ABP&space;=&space;90^\circ&space;$$,&space;$$&space;AB&space;$$&space;is&space;tangent&space;to&space;the&space;circle&space;at&space;$$&space;P&space;$$.$

$\fn_cm&space;Now,&space;let's&space;consider&space;the&space;triangle&space;$$&space;APC&space;$$.&space;Since&space;$$&space;AB&space;$$&space;is&space;tangent&space;to&space;the&space;circle&space;at&space;$$&space;P&space;$$,&space;$$&space;\angle&space;PAB&space;=&space;\angle&space;PAC&space;$$&space;(angles&space;between&space;tangent&space;and&space;radius&space;are&space;equal).&space;Since&space;$$&space;\angle&space;ABC&space;=&space;\angle&space;APC&space;$$,&space;$$&space;\triangle&space;ABC&space;$$&space;is&space;similar&space;to&space;$$&space;\triangle&space;APC&space;$$&space;(by&space;Angle-Angle&space;similarity&space;criterion).$

$\fn_cm&space;Therefore,&space;the&space;ratio&space;of&space;the&space;sides&space;opposite&space;to&space;the&space;equal&space;angles&space;is&space;equal.&space;That&space;is:&space;$&space;\frac{BC}{AC}&space;=&space;\frac{AC}{AB}&space;$&space;$&space;BC&space;\cdot&space;AB&space;=&space;AC^2&space;$&space;But&space;$$&space;AB&space;$$&space;is&space;the&space;diameter&space;of&space;the&space;circle,&space;so&space;$$&space;AB&space;=&space;2&space;\cdot&space;OP&space;$$,&space;where&space;$$&space;OP&space;$$&space;is&space;the&space;radius&space;of&space;the&space;circle.$

$\fn_cm&space;So,&space;we&space;have:&space;$&space;BC&space;\cdot&space;2&space;\cdot&space;OP&space;=&space;AC^2&space;$&space;$&space;BC&space;=&space;\frac{AC^2}{2&space;\cdot&space;OP}&space;$&space;Since&space;$$&space;OP&space;$$&space;is&space;the&space;radius&space;of&space;the&space;circle,&space;and&space;$$&space;\triangle&space;ABC&space;$$&space;is&space;a&space;right&space;triangle,&space;$$&space;OP&space;=&space;OM&space;$$,&space;where&space;$$&space;M&space;$$&space;is&space;the&space;midpoint&space;of&space;$$&space;BC&space;$$.&space;Therefore:&space;$&space;BC&space;=&space;\frac{AC^2}{2&space;\cdot&space;OM}&space;$&space;$&space;2&space;\cdot&space;BC&space;=&space;\frac{AC^2}{OM}&space;$&space;$&space;2&space;\cdot&space;BC&space;=&space;AC&space;\cdot&space;\frac{AC}{OM}&space;$&space;$&space;2&space;\cdot&space;BC&space;=&space;AC&space;\cdot&space;MC&space;$$

$\fn_cm&space;So,&space;we've&space;shown&space;that&space;$$&space;2&space;\cdot&space;BC&space;=&space;AC&space;\cdot&space;MC&space;$$,&space;which&space;implies&space;that&space;$$&space;BM&space;=&space;MC&space;$$.&space;Therefore,&space;the&space;tangent&space;to&space;the&space;circle&space;at&space;$$&space;P&space;$$&space;bisects&space;$$&space;BC&space;$$.$

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To prove that quadrilateral $\fn_cm&space;$$AOBP$$&space;is&space;cyclic,&space;we&space;need&space;to&space;show&space;that&space;its&space;opposite&space;angles&space;are&space;supplementary,&space;i.e.,&space;$$\angle&space;AOB&space;+&space;\angle&space;APB&space;=&space;180^\circ$$.&space;Let's&space;denote:&space;-&space;$$O$$&space;as&space;the&space;center&space;of&space;the&space;circle,&space;-&space;$$P$$&space;as&space;the&space;external&space;point&space;from&space;which&space;tangents&space;are&space;drawn,&space;-&space;$$A$$&space;and&space;$$B$$&space;as&space;the&space;points&space;of&space;tangency&space;of&space;the&space;tangents&space;with&space;the&space;circle.&space;Since&space;$$PA$$&space;and&space;$$PB$$&space;are&space;tangents&space;drawn&space;from&space;an&space;external&space;point&space;$$P$$&space;to&space;the&space;circle,&space;they&space;are&space;equal&space;in&space;length.&space;So,&space;$$PA&space;=&space;PB$$.&space;Now,&space;let's&space;consider&space;triangles&space;$$PAO$$&space;and&space;$$PBO$$:$

$\fn_cm&space;1.&space;In&space;triangle&space;$$PAO$$,&space;$$PO$$&space;is&space;the&space;radius&space;of&space;the&space;circle&space;and&space;$$PA$$&space;is&space;a&space;tangent&space;to&space;the&space;circle,&space;so&space;$$PO&space;\perp&space;OA$$.&space;2.&space;Similarly,&space;in&space;triangle&space;$$PBO$$,&space;$$PO&space;\perp&space;OB$$.&space;Therefore,&space;triangles&space;$$PAO$$&space;and&space;$$PBO$$&space;are&space;both&space;right&space;triangles&space;with&space;the&space;hypotenuse&space;$$PO$$.&space;Since&space;$$PA&space;=&space;PB$$&space;(both&space;are&space;tangents&space;to&space;the&space;circle&space;from&space;point&space;$$P$$)&space;and&space;$$PO$$&space;is&space;the&space;common&space;side,&space;triangles&space;$$PAO$$&space;and&space;$$PBO$$&space;are&space;congruent&space;by&space;the&space;Hypotenuse-Leg&space;(HL)&space;criterion&space;of&space;congruence.&space;As&space;a&space;result,&space;$$\angle&space;APO&space;=&space;\angle&space;BPO$$,&space;because&space;corresponding&space;parts&space;of&space;congruent&space;triangles&space;are&space;congruent.$

$\fn_cm&space;Now,&space;in&space;cyclic&space;quadrilateral&space;$$AOBP$$,&space;opposite&space;angles&space;are&space;supplementary.&space;So,&space;$\angle&space;AOB&space;+&space;\angle&space;APB&space;=&space;180^\circ$&space;Since&space;$$\angle&space;APO&space;=&space;\angle&space;BPO$$,&space;we&space;can&space;rewrite&space;the&space;equation&space;as:&space;$\angle&space;AOB&space;+&space;\angle&space;APO&space;=&space;180^\circ$&space;$\angle&space;AOB&space;+&space;\angle&space;BPO&space;=&space;180^\circ$&space;Since&space;$$\angle&space;AOB$$&space;and&space;$$\angle&space;BPO$$&space;are&space;angles&space;in&space;a&space;cyclic&space;quadrilateral,&space;they&space;add&space;up&space;to&space;$$180^\circ$$.&space;Therefore,&space;quadrilateral&space;$$AOBP$$&space;is&space;a&space;cyclic&space;quadrilateral.$

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