Question

Through a point T, a tangent TA and a secant TPQ are drawn to a circle AQP. If the chord AB is drawn parallel to PQ, prove that the triangles PAT and BAQ are similar.

Answer 1

Given: AB is parallel to PQ and AT is a tangent.

According to the alternate segment theorem, angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

$⟹\angle TAP=\angle AQP-\left(1\right)$

Since $AB\parallel PQ$

$\angle PQA=\angle QAB–\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\angle TAP=\angle BAB$

Also ABQP is a cyclic quadrilateral. So opposite angles are supplementary.

$\angle QPA+\angle QBA=180⟹\angle QBA=180-\angle QPA$

$\angle QPA+\angle APT=180$

$⟹\angle APT=180-\angle QPA$

$\angle APT=\angle QBA$

In triangles PAT and BAQ

The triangles are similar.