Question

In figure, OA OD, OC OB, OD = OA and OC = OB.

Prove that AB = CD​

Answer 1

Given:

$\sf{\angle DOA = \angle BOC = 90°}$

$\sf{OD = OA}$

$\sf{OC = OB}$

Proof:

Let $\sf{\angle AOC = x}$

From ∆DOC and ∆AOB,

$\sf{•\:\angle DOC = \angle AOB = 90° + x}$

$\sf{•\:OD = OA}$

$\sf{•\:OC = OB}$

By SAS Congruency,

$\bold{∆DOC \cong ∆AOB}$

By corresponding parts of congruent triangles (CPCT),

$\bold{\red{AB=CD}}$

★ Hence proved