Prove that the diagonals of a trapezium divide each other proportionally.

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Solution

Given :
A trapezium ABCD in which $A B ∥ D C$ and its diagonals AC and BD intersect at O.
To prove :
$\frac{A O}{O C} = \frac{B O}{O D}$

Construction :
Through O, draw $E O ∥ A B$ , meeting AD at E.
Proof :
In $△ A D C, E O ∥ D C$
Therefore,
$\frac{A E}{E D} = \frac{A O}{O C}$ .........(1)

In $△ D A B, E O ∥ A B$
Therefore,
$\frac{A E}{E D} = \frac{B O}{O D}$ ..........(2)

From 1 and 2, we get,
$\frac{A O}{O C} = \frac{B O}{O D}$

Hence, proved.

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Similar questions

The diagonals of a quadrilateral ABCD interest each other such that ao/no=co/do prove that its a trapezium

A C

B D

A B C D

A B ∥ D C

O .

a r (A O D) = a r (B O C)

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

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