Question

$OA=OB,OC=OD$, $\angle AOB=\angle COD$. Prove that $AC=BD$

Answer 1

Given $\angle AOB=\angle COD\angle AOB-\angle COB=\angle COD-\angle COB\therefore \angle AOC=\angle DOB$

In $\Delta AOC$ and $\Delta DOB$

$i)OA=OB$(Given)

$ii)OC=\angle OD$ (Given)

$iii)\angle AOC=\angle DOB$ (Proved above)

$\therefore \Delta AOC\cong \angle DOB$

(SAS Axiom)

$\therefore AC=BD$