Name: Proof of Angle in the Same Segment of a Circle
Uploaded: 2019-07-19
Duration: 1 min 26 s
Description: The most important and simple theorem from the chapter circle that angle in the same segment of the circle are equal. Let's derive this theorem.

Question

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Accepted Answer

Given: $ABCD$ is a cyclic quadrilateral $AC, BD$ are diameters of the circle.

To prove: $ABCD$ is a rectangle.

Proof: Let $A B C D$ a cyclic quadrilateral having diagonals as $B D$ and $A C$ intersecting each other at point $O .$

Now, Consider $B D$ as a chord.

We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.

$∴ ∠ B A D = \frac{1}{2} ∠ B O D$ $= \frac{1}{2} (180 °)$ $= 90 °$

$∠ B C D + ∠ B A D = 180 °$ (Cyclic quadrilateral)

$∠ B C D = 180 ° - 90 ° = 90 °$

Similarly, $∠ A D C = 90 °$

$∠ A D C + ∠ A B C = 180 °$ (Cyclic quadrilateral)

$90 ° + ∠ A B C = 180 °$

$∠ A B C = 90 °$

Since, each interior angle of cyclic quadrilateral is of $90 ° .$

From $△ AOD$ and $△ BOC$

$AO = OC$ (radius)

$BO = OD$ (radius)

$∠ AOD = ∠ BOC$ (vertically opposite angles)

$∴ △ AOD ≅ △ BOC$ (by $S A S$ rule of congruence)

$\Rightarrow AD = BC$ and $∠ OAD = ∠ OCB$ (by Corresponding sides and corresponding angles of a congruence triangles)

$∴ AD ∥ BC$ (alternate interior angles)

$∴ ABCD$ is a parallelogram.

We know that in a parallelogram, if one angle is $90 °$ then, it is rectangle.

$∠ ABC = 90 °$

Hence, it is a rectangle.

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Important Questions on Circles

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