Question

PQ is a diameter of a circle and AB is a perpendicular chord cutting it at N. Prove that PN is equal in length to the perpendicular from P on to the tangent at A.

Answer 1

Given: $AB\perp PQ$

$AX$ is a tangent at $A,AX\perp PX$

$PQ$ is a diameter

In $△PAQ$

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

Since $AX$ 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.

$\angle XAP=\angle AQP–\left(1\right)$

$\angle APQ+\angle PQA+\angle QAP=180$

$\angle APQ=90-\angle PQA$

$\angle APQ=90-\angle XAP–\left(2\right)$

In $△AXP$

$\angle PXA=180–90-\angle XAP$

$\angle PXA=90-\angle XAP$

From $\left(2\right)$ and $\left(3\right)$

$\angle APQ=\angle PXA$

In triangles $AXPandPXA$

$\angle APQ=\angle PXA$

$\angle AXP=\angle ANP=90$

$AP=AP$ (Common)

By RHS congruency $△AXP\cong △ANP$

By CPCT,

$PN=PX$