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Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.
In circle I,
For chord AP,
∠PBA = ∠ACP (Angles in the same segment are equal) — (i)
In circle II,
For chord DQ,
∠DBQ = ∠QCD (Angles in same segment) — (ii)
ABD and PBQ are line segments intersecting at B.
∠PBA = ∠DBQ (Vertically opposite angles) —iii
From the equations (i), (ii) and (iii),
∠ACP = ∠QCD
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If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Two circles are drawn with sides AB and AC of the triangle ΔABC as diameters. Both
these circles intersect each other at D.
To Prove:
D lies on
BC
Construction:
Join AD
To prove:
Since,AC
and AB are the diameters of the two circles.
∠ADB =90°......(i)
∠ADC =
90°......(ii)
(Angle in the semi circle)
On adding eq i & ii
∠ADB + ∠ADC = 180°
∠BDC=
180°
Hence,BDC is straight line.
hence, our assumption was wrong.
So , point of intersection D
lies on the third side.
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ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
In ΔABC,
∠ABC + ∠BCA + ∠CAB = 180° (Angle sum property of a triangle)
⇒ 90° + ∠BCA + ∠CAB = 180°
⇒ ∠BCA + ∠CAB = 90° ... (1)
In ΔADC,
∠CDA + ∠ACD + ∠DAC = 180° (Angle sum property of a triangle)
⇒ 90° + ∠ACD + ∠DAC = 180°
⇒ ∠ACD + ∠DAC = 90° ... (2)
Adding equations (1) and (2), we obtain
∠BCA + ∠CAB + ∠ACD + ∠DAC = 180°
⇒ (∠BCA + ∠ACD) + (∠CAB + ∠DAC) = 180°
∠BCD + ∠DAB = 180° ... (3)
However, it is given that
∠B + ∠D = 90° + 90° = 180° ... (4)
From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180°. Therefore, it is a cyclic quadrilateral.
Consider chord CD.
∠CAD = ∠CBD (Angles in the same segment)
Prove that a cyclic parallelogram is a rectangle.
Let ABCD be a cyclic parallelogram.
∠A + ∠C = 180° (Opposite angles of a cyclic quadrilateral) ... (1)
We know that opposite angles of a parallelogram are equal.
∴ ∠A = ∠C and ∠B = ∠D
From equation (1),
∠A + ∠C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2 ∠A = 180°
⇒ ∠A = 90°
Parallelogram ABCD has one of its interior angles as 90°. Therefore, it is a rectangle.
In Figure, A,B and C are three points on a circle with centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other
than the arc ABC, find ∠ADC.
It can be observed that
∠AOC = ∠AOB + ∠BOC
= 60° + 30°
= 90°
We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at
a point on the minor arc and also at a point on the major arc.
In ΔOAB,
AB = OA = OB = radius
∴ ΔOAB is an equilateral triangle.
Therefore, each interior angle of this triangle will be of 60°.
∴ ∠AOB = 60°
In cyclic quadrilateral ACBD,
∠ACB + ∠ADB = 180° (Opposite angle in cyclic quadrilateral)
⇒ ∠ADB = 180° − 30° = 150°
Therefore, angle subtended by this chord at a point on the major arc and the minor arc are 30° and 150° respectively.
In Figure, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR.
Consider PR as a chord of the circle.
Take any point S on the major arc of the circle.
PQRS is a cyclic quadrilateral.
∠PQR + ∠PSR = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠PSR = 180° − 100° = 80°
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠POR = 2∠PSR = 2 (80°) = 160°
In ΔPOR,
OP = OR (Radii of the same circle)
∴ ∠OPR = ∠ORP (Angles opposite to equal sides of a triangle)
∠OPR + ∠ORP + ∠POR = 180° (Angle sum property of a triangle)
2 ∠OPR + 160° = 180°
2 ∠OPR = 180° − 160° = 20º
∠OPR = 10°
In Figure, ∠ ABC = 69°, ∠ ACB = 31°, find ∠ BDC.
In ΔABC,
∠BAC + ∠ABC + ∠ACB = 180° (Angle sum property of a triangle)
⇒ ∠BAC + 69° + 31° = 180°
⇒ ∠BAC = 180° − 100º
⇒ ∠BAC = 80°
∠BDC = ∠BAC = 80° (Angles in the same segment of a circle are equal)
In Figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.
In ΔCDE,
∠CDE + ∠DCE = ∠CEB (Exterior angle)
⇒ ∠CDE + 20° = 130°
⇒ ∠CDE = 110°
However, ∠BAC = ∠CDE (Angles in the same segment of a circle)
⇒ ∠BAC = 110°
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°,
∠ BAC is 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD.
For chord CD,
∠CBD = ∠CAD (Angles in the same segment)
∠CAD = 70°
∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
∠BCD + ∠BAD = 180° (Opposite angles of a cyclic quadrilateral)
∠BCD + 100° = 180°
∠BCD = 80°
In ΔABC,
AB = BC (Given)
∴ ∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)
⇒ ∠BCA = 30°
We have, ∠BCD = 80°
⇒ ∠BCA + ∠ACD = 80°
30° + ∠ACD = 80°
⇒ ∠ACD = 50°
⇒ ∠ECD = 50°
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.
(Consider BD as a chord)
∠BCD + ∠BAD = 180° (Cyclic quadrilateral)
∠BCD = 180° − 90° = 90°
(Considering AC as a chord)
∠ADC + ∠ABC = 180° (Cyclic quadrilateral)
90° + ∠ABC = 180°
∠ABC = 90°
Each interior angle of a cyclic quadrilateral is of 90°. Hence, it is a rectangle.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Consider a trapezium ABCD with AB | |CD and BC = AD.
Draw AM ⊥ CD and BN ⊥ CD.
In ΔAMD and ΔBNC,
AD = BC (Given)
∠AMD = ∠BNC (By construction, each is 90°)
AM = BN (Perpendicular distance between two parallel lines is same)
∴ ΔAMD ≅ ΔBNC (RHS congruence rule)
∴ ∠ADC = ∠BCD (CPCT) ... (1)
∠BAD and ∠ADC are on the same side of transversal AD.
∠BAD + ∠ADC = 180° ... (2)
∠BAD + ∠BCD = 180° [Using equation (1)]
This equation shows that the opposite angles are supplementary.
Therefore, ABCD is a cyclic quadrilateral.
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