Question

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

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Show that `(OA)/(OC) = (OB)/(OD)`.

Answer 1

We have,

ABCD is a trapezium with AB || DC

In ΔAOB and ΔCOD

∠AOB = ∠COD                   ...[Vertically opposite angles]

∠OAB = ∠OCD                   ...[Alternate interior angles]

Then, ΔAOB ~ ΔCOD         ...[By AA similarity]

`therefore"OA"/"OC"="OB"/"OD"`                ...[Corresponding parts of similar Δ are proportional.]