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

Asked by Pragya Singh | 1 year ago |  64

Solution :-

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.

$$\dfrac{DO}{BO}=\dfrac{OC}{OA}$$

$$\dfrac{OA}{OC}=\dfrac{OB}{OD}$$

Hence, proved.

Answered by Abhisek | 1 year ago

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