Question

In the given figure PQ is a tangent to the circle at A. AB and AD are bisectors of $\angle$CAQ and $\angle$PAC. If $\angle$BAQ$={30}^o$, prove that.
(I) BD is a diameter of the circle.
(II) ABC is an isosceles triangle.

Answer 1

$\angle BAQ={30}^{\circ }$

$\text{Since AB is the bisector of}\angle CAQ$

$\angle CAB=\angle BAQ={30}^{\circ }$

$\angle CAQ=\angle CAB+\angle BAQ={60}^{\circ }$

$\text{We can also see that:}\angle PAC+\angle CAQ={180}^{\circ }$

$So,\angle PAC={120}^{\circ }$

$\text{Since AD bisects}\angle PAC$

$So,\angle PAD=\angle DAC=\frac{\angle PAC}{2}={60}^{\circ }$

$So,\angle DAB={90}^{\circ }$

$\text{And we know that only diameter subtends an angle of}{90}^{\circ }\text{on the circle}$

$\text{Hence part (I) is proved}$

$\text{Now we see that}\angle CAB\text{\&}\angle ACB\text{are equal because they share a same side in front on them}$

$\text{Finally we can see two angles same in}\Delta ABC\text{So it is an Isosceles triangle, Hence (II) is proved}$