Question

If $△ABC$ is isosceles with AB = AC, prove that the tangent at A to the circumcircle of $△ABC$ is parallel to BC. [2 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark

As AB = AC

Let $\angle ABC=\angle ACB=x$

$\therefore \angle AOB=2\angle ACB=2x$

In $△AOB$

$\angle OBA=\angle OAB$

as OA = OB (radius)

Now

$\angle OBA+\angle OAB+\angle AOB={180}^{\circ }$ (By Angle Sum Property)

$\angle OAB+\angle OAB+2x={180}^{\circ }$

$\angle OAB+x={90}^{\circ }$

$\angle OAB={90}^{\circ }-x$

Now

$\because$ , $\angle TAO={90}^{\circ }$

$\Rightarrow$ $\angle TAB=\angle TAO-\angle BAO$

$\Rightarrow$$\angle TAB={90}^{\circ }-({90}^{\circ }-x)$

$\angle TAB=x=\angle ABC$

Therefore TA is parallel to BC

Hence proved.