Question

A tangent PT is drawn parallel to a chord AB as shown in figure. Prove that APB is an isosceles triangle.

Answer 1

$\angle$TPA = $\angle$PAB …..(i) (Since TP $\parallel$ AB)

We know $\angle$POA = 2$\angle$PBA because the angle subtended by the chord at the center is twice the angle subtended on the circle

In $△$POA

OP = OA (radius)

Therefore $\angle$OAP = $\angle$OPA

$\angle$OAP + $\angle$OPA + $\angle$POA = ${180}^{\circ }$

$\angle$OPA + $\angle$OPA + 2$\angle$PBA = ${180}^{\circ }$

2$\angle$OPA + 2$\angle$PBA = ${180}^{\circ }$

$\angle$OPA + $\angle$PBA = ${90}^{\circ }$ …..(ii)

As the tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\angle$OPT = ${90}^{\circ }$

$\angle$OPA + $\angle$APT = ${90}^{\circ }$ …..(iii)

Comparing (ii) and (iii)

$\angle$PBA = $\angle$APT ….(iv)

Comparing (i) and (iv)

$\angle$PBA = $\angle$PAB

$\Rightarrow$PA = PB

$\Rightarrow$$△$PAB is isosceles