Question

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

Answer 1

Given: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O

To Prove: $\frac{OA}{OC}=\frac{OB}{OD}$

Proof:
In ∆DOC and ∆BOA,
$\angle$CDO = $\angle$ABO [Alternate interior angles as AB || CD]
$\angle$DCO = $\angle$BAO [Alternate interior angles as AB || CD]
$\angle$DOC = $\angle$BOA [Vertically opposite angles]
∴ ∆DOC ∼ ∆BOA [AAA similarity criterion]

∴ $\frac{DO}{BO}=\frac{OC}{OA}$ [coresponding sides are proportional]

⇒ $\frac{OA}{OC}=\frac{OB}{OD}$

Hence Proved