Question

22. If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.
2​

Answer 1

Given :-

Let ABCD be a quadrilateral in which diagonals AC and BD intersects at O and

$\sf \: \dfrac{OA}{OC} = \dfrac{OB}{OD}$

$\rm :\implies\:\dfrac{OA}{OB} = \dfrac{OC}{OD}$

To prove :-

• ABCD is a trapezium

Concept Used :-

• 1. Similarity of triangles

• 2. If one pair of opposite sides are parallel, then quadrilateral is a trapezium.

Proof :-

Consider,

$\rm :\longmapsto\:In \: \triangle \: AOB \: and \:\triangle \: COD$

$\rm :\longmapsto\:\:\dfrac{OA}{OB} = \dfrac{OC}{OD} \: \: \: \: (given)$

$\rm :\longmapsto\: \angle \: AOB \: = \:\angle \: COD \: \: \: (vertically \: opp. \: angles)$

$\rm :\longmapsto\: \: \triangle \: AOB \: \sim\:\triangle \: COD \: \: \: \: (SAS)$

$\bf\implies \: \angle \: ABO \: = \: \angle \: ODC \: \: \: \: \: (CPST)$

But these are alternate interior angles

$\bf\implies \:AB \: \parallel \: CD$

$\bf\implies \:ABCD \: is \: a \: trapezium.$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

Basic Proportionality theorem :-

• It states that if a line is drawn parallel to one side of a triangle to intersect other two sides in two distinct points, it divides the other two sides in same ratio.

Pythagoras theorem :-

• It states that in a right angle triangle, the square of the hypotenuse is equal to sum of the squares of remaining two sides.

Converse of Pythagoras Theorem :-

• It states that if in a triangle, the square of longest side is equal to sum of the squares of remaining two sides, then angle opposite to longer side is 90°.