Question

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium. ​

Answer 1

SOLUTION:

Given:

ABCD is a quadrilateral. Its diagonal AC and BD intersect each other at O such that

$\frac{AO}{BO}= \frac{CO}{DO} \implies \boxed{\frac{AO}{\bf CO}=\frac{\bf BO}{DO}}$  ......(i)

To prove;

ABCD is a trapezium

Construction:

Through O, we can draw OE║BA

Let OE meets AD at E.

In ΔDAB

We have EO║AB      [∵ by construction]

$\implies \frac{DE}{AE}=\frac{DO}{BO}$

[∴ By Basic Proportionality Theorem - Thales theorem, If a line is drawn parallel to one side of a triangle to intersect the other two sides in the distinct points, then the other two sides are divided in the same ratio.]

or also

$\implies \frac{AE}{DE}=\frac{BO}{DO}$

[By taking the reciprocals]   ........(ii)

Now, from equations (i) and (ii), we get

$\frac{AO}{CO} = \frac{AE}{DE} \implies \boxed{OE \parallel CD}$

[∵ By Converse of Basic Proportionality Theorem - If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side]

So,

BA ║OE and OE║CD

∴ BA║CD  [As lines are parallel to the same line are also parallel t each other]

Hence,

Quadrilateral ABCD is a trapezium. (Proved)

Answer 2

Step-by-step explanation:

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