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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.


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Solution

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

In OAC and ODB

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

Similarly

ABD=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.


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