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Answered on 16 Apr Learn Constructions

Sadika

To construct a triangle read more

To construct a triangle

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Answers 1 Comments Answered on 16 Apr Learn Constructions

Sadika

To construct the tangents from point read more

To construct the tangents from point

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Answers 1 Comments Answered on 16 Apr Learn Triangles

Sadika

To prove that the above equation read more

To prove that the above equation

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Answered on 16 Apr Learn Triangles

Sadika

Let's denote the area of triangle read more

Let's denote the area of triangle

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Answers 1 Comments Answered on 16 Apr Learn Triangles

Sadika

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.

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Answers 1 Comments Answered on 16 Apr Learn Triangles

Sadika

Given :: solution:: for above Question read more

Given :: solution:: for above Question

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Answered on 16 Apr Learn Triangles

Sadika

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.

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Answers 1 Comments Answered on 16 Apr Learn Circles

Sadika

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:

- 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 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

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Answers 1 Comments Answered on 16 Apr Learn Circles

Sadika

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.

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Answered on 16 Apr Learn Circles

Sadika

To prove that quadrilateral read more

To prove that quadrilateral

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