Question

In the adjacent figure, AB is a chord of circle with centre O. CD is the diameter perpendicular to AB. Show that AD = BD.

Answer 1

Since $CD$ is perpendicular to $AB$ and passes through center $O$, therefore, $CD$ bisects $AB$

$⟹AE=BE$ where E is the point of intersection of $AB$ and $CD$

In $△ADE$ and $△BDE$

$ED=ED$ [common side]

$\angle DEA=\angle DEB={90}^{\circ }$

$AE=BE$ [Proved above]

By $SAS$ congruency the 2 triangles $△ADE\cong △BDE$

$\therefore AD=BD$ [By CPCT]