AB is a diameter and AC is a chord of a circle with centre O such that BAC = . The tangent at C intersects extended AB at a point D. Prove that BC = BD.

Name: AB is a diameter and AC is a chord of a circle with centre O such that angle BAC = 30°degree The tangent at C intersects extended AB at a point D. Prove that BC = BD.
Uploaded: Sat, 2021-01-30 16:13
Description: AB is a diameter and AC is a chord of a circle with centre O such that angle BAC = 30°degree The tangent at C intersects extended AB at a point D. Prove that BC = BD.

AB is a diameter and AC is a chord of a circle with centre O such that angle BAC = 30°degree The tangent at C intersects extended AB at a point D. Prove that BC = BD.

AB is a diameter and AC is a chord of a circle with centre O such that $\angle$ BAC = $30^{\circ}$ . The tangent at C intersects extended AB at a point D. Prove that BC = BD.

Answers (1)

Solution

Given AB is a diameter and AC is a chord of circle with center O.
$\angle BAC= 30^{\circ}$
To Prove : BC = BD
Construction : Join B and C
Proof : $\angle BCD= \angle CAB$ [Angle is alternate segment]
$\angle BCD= 30^{\circ}\: \left [ Given \right ]$
$\angle BCD= 30^{\circ}\cdots \left ( i \right )$
$\angle ACB= 90^{\circ}$ [Angle in semi circle formed is 90°]
In $\bigtriangleup$ ABC
$\angle CAB+\angle ABC+\angle BCA= 180^{\circ}$ [Sum of interior angles of a triangle is 180°]
$30^{\circ}+\angle ABC+90^{\circ}= 180^{\circ}$
$\angle ABC= 180^{\circ}-90^{\circ}-30^{\circ}$
$\angle ABC= 60^{\circ}$
Also , $\angle ABC+\angle CBD= 180^{\circ}$ [Linear pair]
$\angle CBD= 180^{\circ}-60^{\circ}$
$\angle CBD= 120^{\circ}$
$\angle ABC= 60^{\circ}$
In $\bigtriangleup CBD$
$\angle CBD+\angle BDC+\angle DCB= 180^{\circ}$
$\angle BDC= 30^{\circ}\cdots \left ( ii \right )$
From equation (i) and (ii)
$\angle BCD= \angle BDC$
$\Rightarrow BC= BD$ [ $\because$ Sides opposite to equal angles are equal]
Hence Proved