Question

Given a parallelogram ABCD. Prove that $OC=OA$

Answer 1

Given:- $ABCD$ is a parallelogram with $AC\&BD$ as diagonals.

$ABCD$ is a parallelogram.

$\therefore AD\parallel BC⟹\angle BDA=\angle CBD$ and $\angle DAC=\angle ACB$ ...... (alternate angles)

$or\angle ODA=\angle CBO$ and $\angle DAO=\angle OCB........\left(i\right)$

Now between $\Delta ADO$ and $\Delta COB$ we have $AD=BC$ ...... (opposite side of a parallelogram)

$\angle ODA=\angle CBO$ and $\angle DAO=\angle OCB$ ..... by $\left(i\right)$

$\therefore$ By ASA test $\Delta AOD$ and $\Delta COB$ are congruent.

$\therefore OC=OA$