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In ΔDOC and ΔBOA,
- ∠CDO = ∠ABO (Alternate interior angles as AB || CD)
- ∠DCO = ∠BAO (Alternate interior angles as AB || CD)
- ∠DOC = ∠BOA (Vertically opposite angles)
Therefore,
ΔDOC ∼ ΔBOA [ BY AAA similarity]
Implies,
OA/OC = OB/OD (Corresponding sides of similar triangles are proportional)
or,
AO/OC = BO/DO
[Hence: Proved] Q.E.D
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