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Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion for two triangles, show that OAOCOBODOAOC=OBOD - Mathematics

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Questions

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

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Show that `"OA"/"OC"="OB"/"OD"`.

Sum
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Solution

We have,

ABCD is a trapezium with AB || DC

In ΔAOB and ΔCOD

∠AOB = ∠COD                   ...[Vertically opposite angles]

∠OAB = ∠OCD                   ...[Alternate interior angles]

Then, ΔAOB ~ ΔCOD         ...[By AA similarity]

`therefore"OA"/"OC"="OB"/"OD"`                ...[Corresponding parts of similar Δ are proportional.]

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Chapter 7: Triangles - Exercise 7.5 [Page 74]
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