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# 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 AO/OC = OB/OD Q.3

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For class 10 Triangles chapter of exercise 6.3 How i solve this question in easy way because i don’t know how to solve this problem 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 AO/OC = OB/OD

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### 2 Answers

1. In Î”DOC and Î”BOA,

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

âˆ´âˆ CDO = âˆ ABO

Similarly,

âˆ DCO = âˆ BAO

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

âˆ´âˆ DOC = âˆ BOA

Hence, by AAA similarity criterion,

Î”DOC ~ Î”BOA

Thus, the corresponding sides are proportional.

DO/BO = OC/OA

â‡’OA/OC = OB/OD

Hence, proved.

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2. In Î”DOC and Î”BOA,

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

âˆ´âˆ CDO = âˆ ABO

Similarly,

âˆ DCO = âˆ BAO

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

âˆ´âˆ DOC = âˆ BOA

Hence, by AAA similarity criterion,

Î”DOC ~ Î”BOA

Thus, the corresponding sides are proportional.

DO/BO = OC/OA

â‡’OA/OC = OB/OD

Hence, proved.

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