Question

Question 8
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.

Answer 1

A circle is drawn with centre O and AB is a diameter.
AC is a chord such that $\angle BAC={30}^{\circ }$
Given: AB is the diameter and AC is a chord of circle with center O, $\angle BAC={30}^{\circ }$
To Prove: BC =BD

Proof: $\angle BCD=\angle CAB$ [ alternate segment theorem]
$\angle CAB={30}^{\circ }$ [given]
$\therefore \angle BCD={30}^{\circ }$ (i)
$\angle ACB={90}^{\circ }$ [ angle in semi-circle is right angle]
In $\Delta$ ABC.
$\angle A+\angle B+\angle C={180}^{\circ }$
${30}^{\circ }+\angle B+{90}^{\circ }={180}^{\circ }$
$\angle B={60}^{\circ }$
$\Rightarrow \angle CBA+\angle CBD={180}^{\circ }$
[linear pair]
Also $\angle CBD={180}^{\circ }-{60}^{\circ }-{120}^{\circ }$ $[\because \angle CBA={60}^{\circ }]$
Now, in $\Delta$ CBD
$\angle CBD+\angle BDC+\angle DCB={180}^{\circ }$
$\Rightarrow {120}^{\circ }+\angle BDC+{30}^{\circ }={180}^{\circ }$
$\Rightarrow \angle BDC={30}^{\circ }$ ...(ii)
From Eqs. (i) and (ii)
$\angle BCD=\angle BDC$
$\therefore$ BC = BD [ sides opposite to equal angles are equal.]