Question

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC= OB/OD.​

Answer 1

Given:-

• Diagonals AC & BD.
• AB || DC intersect each other at point O.

To ShOW:-

• $\implies\frac{OA}{OC} \:=\:\frac{OB}{OD}$

SoluTion:-

• In ΔAOB and ΔCOD,

1. ∠AOB  = ∠COD  (Vertically opposite angles are equal)
2. ∠OAB   = ∠OCD (Alternate interior angles are equal)

By Similar criterion,

ΔAOB∼ΔCOD

∴ OA/OC = OB/OD

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