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State “Basic Proportionality Theorem” and use it to prove the following :
In a quadrilateral ABCD, diagonals AC and BD intersect each other at O such that $\frac{AO}{BO} = \frac{CO}{DO}$ as shown in the given figure. Prove that ABCD is a trapezium.
In a quadrilateral ABCD, diagonals AC and BD intersect each other at O such that $\frac{AO}{BO} = \frac{CO}{DO}$ as shown in the given figure. Prove that ABCD is a trapezium.
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Solution: Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given : ABCD is a quadrilateral in which $\frac{AO}{BO} = \frac{CO}{DO}$
To Prove : $AB \parallel CD$
Construction : Draw $OE \parallel AB$
Proof : In $\Delta DAB, OE \parallel AB$
$\therefore \frac{DE}{AE} = \frac{DO}{BO}$ (by BPT)
Also $\frac{AO}{BO} = \frac{CO}{DO}$ (given)
$\Rightarrow \frac{DO}{BO} = \frac{CO}{AO}$
$\therefore \frac{DE}{AE} = \frac{CO}{AO}$
In $\Delta ADC, \frac{DE}{AE} = \frac{CO}{AO}$
$\therefore OE \parallel CD$ (by converse of BPT)
As $OE \parallel AB$ and $OE \parallel CD$
$\therefore AB \parallel CD$
Hence, ABCD is a trapezium
Given : ABCD is a quadrilateral in which $\frac{AO}{BO} = \frac{CO}{DO}$
To Prove : $AB \parallel CD$
Construction : Draw $OE \parallel AB$
Proof : In $\Delta DAB, OE \parallel AB$
$\therefore \frac{DE}{AE} = \frac{DO}{BO}$ (by BPT)
Also $\frac{AO}{BO} = \frac{CO}{DO}$ (given)
$\Rightarrow \frac{DO}{BO} = \frac{CO}{AO}$
$\therefore \frac{DE}{AE} = \frac{CO}{AO}$
In $\Delta ADC, \frac{DE}{AE} = \frac{CO}{AO}$
$\therefore OE \parallel CD$ (by converse of BPT)
As $OE \parallel AB$ and $OE \parallel CD$
$\therefore AB \parallel CD$
Hence, ABCD is a trapezium