If in the figure, O is the point of intersect | vedclass

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(A) In $	riangle OAC$ and $	riangle ODB$:$1$. $\angle AOC = \angle DOB$ (Vertically opposite angles).$2$. $\angle OAC = \angle ODB$ (Angles subtende

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isosceles and similar

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(A) In $	riangle OAC$ and $	riangle ODB$:$1$. $\angle AOC = \angle DOB$ (Vertically opposite angles).$2$. $\angle OAC = \angle ODB$ (Angles subtended by the same arc $CB$ at the circumference are equal).$3$. $\angle OCA = \angle OBD$ (Angles subtended by the same arc $AD$ at the circumference are equal).Thus, $	riangle OAC \sim 	riangle ODB$ by $AAA$ similarity criterion.Given $OB = OD$, in $	riangle ODB$, the angles opposite to these sides must be equal, i.e., $\angle ODB = \angle OBD$. Since $\angle OAC = \angle ODB$ and $\angle OCA = \angle OBD$, it follows that $\angle OAC = \angle OCA$. Therefore, $	riangle OAC$ is an isosceles triangle with $OA = OC$.Since $	riangle OAC \sim 	riangle ODB$ and $	riangle OAC$ is isosceles, $	riangle ODB$ must also be isosceles.Hence, the triangles are isosceles and similar.

If in the figure, $O$ is the point of intersection of two chords $AB$ and $CD$ such that $OB = OD$, then triangles $OAC$ and $ODB$ are

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