Question

In fig. two equal chords $AB$ and $CD$ of a circle with centre $O$, intersect each other at $E$. Prove that $AD=CB$.

Answer 1

From the figure there are two triangles $\Delta AED$ and $\Delta CEB$

To prove $\Delta AED$ and $\Delta CEB$

From the figure we get$,$

$<AED=<CEB($Vertically opposite angles are equal$)$

$<A=<C($since $<A$ and $<C$ are on the angle subtended on the same arc$)$

Similarly $<D=<B$

All angle are equal. Therefore $\Delta AED\cong \Delta CEB$

To prove $AD=CB$

Since $\Delta AED\cong \Delta CEB$

By$CPCT($Corresponding parts of congruent triangles$)$

$AE=CE,ED=EB$ and $AD=CB$