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