Question

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

Answer 1

Answer:

In △DOC and △BOA,

AB || CD, thus alternate interior angles will be equal,

∠CDO = ∠ABO

Similarly,

∠DCO = ∠BAO

Also, for the two triangles △DOC and △BOA, vertically opposite angles will be equal;

∠DOC = ∠BOA

Hence, by AAA similarity criterion,

△DOC ~ △BOA

Thus, the corresponding sides are proportional.

DO/BO = OC/OA

OA/OC = OB/OD

Hence, proved.