Question

Diagonals ac and bd of a trapezium abcd with ab ll dc intersect each other at the point o. using a similarity criterion for two triangles, show that oa\oc =ob\od

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

In ΔDOC and ΔBOA,

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

∴∠CDO = ∠ABO

Similarly,

∠DCO = ∠BAO

Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;

∴∠DOC = ∠BOA

Hence, by AAA similarity criterion,

ΔDOC ~ ΔBOA

Thus, the corresponding sides are proportional.

DO/BO = OC/OA

⇒OA/OC = OB/OD

Hence proved