Question

If $O$ is the point of intersection of two chords $AB$ and $CD$ of a circle such that $OB=OD$, then prove that triangles $△OAC$ and $△ODB$ are similar.

Answer 1

Given $O$ is the point of intersection of two chords $AB$ and $CD$ of a circle such that $OB=OD$

In $△OAC$ and $△ODB$

$\angle CAB=\angle CDB$ ( Angle on the common arc $BC$ of a circle are equal )

Similarly

$\angle ABD=\angle ACD$ ( Angle on the common arc $AD$ of a circle are equal )

By $AA$ criteria it can be conclude $△OAC$ and $△ODB$ are similar.

Hence, triangles $△OAC$ and $△ODB$ are similar.