Question

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR. Show that
1) ∆ ABC ~ ∆ PQR.​

Answer 1

Given :-

• Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR.

To Prove :-

• ∆ ABC ~ ∆ PQR

Solution :-

Since, AD is median of ∆ ABC

• Therefore, BD = DC

• ⇛ BC = 2BD -------(1)

Again,

PM is median ∆ PQR,

• Therefore, QM = MR

• ⇛ QR = 2QM--------(2)

Now,

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR.

$\rm :\implies\:\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$

$\rm :\implies\:\dfrac{AB}{PQ} = \dfrac{2BD}{2QM} = \dfrac{AD}{PM} \: \: \: \: (using \: (1) \: and \: (2))$

$\rm :\implies\:\dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM}$

$\bf\implies \: \triangle \: ABD \: \sim \: \triangle \: PQM \: \: \: (SSS \: Similarity)$

$\bf\implies \: \angle \: ABC \: = \: \angle \: PQR \: \: \: \: \: \: (CPST)$

Now,

$\rm :\longmapsto\:In \: \triangle \: ABC \: and \: \triangle \: PQR$

$\rm :\longmapsto\:\dfrac{AB}{PQ} = \dfrac{BC}{QR} \: \: \: \: (given)$

$\rm :\longmapsto\:\angle \: ABC \: = \: \angle \: PQR \: \: \: (proved \: above)$

$\bf\implies \:\triangle \: ABC \: \sim \: \triangle \: PQR \: \: (SAS \: \: Similarity)$

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

Additional Information :-

1. Basic Proportionality Theorem :-

• If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

2. Pythagoras Theorem :-

• Pythagoras theorem states that : In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

3. Converse of Pythagoras Theorem :-

• If the square of a side is equal to the sum of the square of the other two sides, then triangle must be right angle triangle.