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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"`.
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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