Question

If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median of another triangle, then the two triangles are similar.

Accepted Answer

Given: Two triangles $△ ABC$ and $△ DEF$ in which $AP$ and $DQ$ are the medians such that
$\frac{AB}{DE} = \frac{AC}{DF} = \frac{AP}{DQ}$
To prove: $∆ ABC ~ ∆ DEF$
Construction: Product $AP$ to $G$ so that $PG = AP$ . Join $CG$ . Also, produce $DQ$ to $H$ so that $QH = DQ$ Join $FH$ .
Proof: In $∆ APB$ and $∆ GPC$ , we have $BP = CP$ $[∵ AP is the median]$
$AP = GP$
[By construction] and, $∠ APB = ∠ CPG IVertically opposite angles]$
So, by $SAS$ -criterion of congruence, we have
$\Rightarrow ΔAPB ≅ ΔGPC$
$\Rightarrow AB = GC$ ...(i)

Again, In $ΔDQE$ and $ΔHQF,$ we have
$\Rightarrow EQ = FQ$ $[∵ DQ is the median]$
$\Rightarrow DQ = HQ$ [By construction]
$\Rightarrow ∠ DQE = ∠ HQF$ [Vertically opposite angles]
So, by $SAS$ -criterion of congruence, we have $ΔDQE ≅ ΔHQF$
By CPCT $\Rightarrow DE = HF$ ...(ii)

Now, $\frac{AB}{DE} = \frac{AC}{DF} = \frac{AP}{DQ}$
$\Rightarrow \frac{GC}{HF} = \frac{AC}{DF} = \frac{AP}{DQ} [∵ AB = GC and DE = HF (From (i) and (ii))]$
$\Rightarrow \frac{GC}{HF} = \frac{AC}{DF} = \frac{2 AP}{2 DQ}$ $[∵ 2 AP = AG and 2 DQ = DH]$
$\Rightarrow ΔAGC ~ ΔDHF$ [By $SSS$ -criterion of similarity]
$\Rightarrow ∠ 1 = ∠ 2$
Similarly, we have
$∠ 3 = ∠ 4$
$∴ ∠ 1 + ∠ 3 = ∠ 2 + ∠ 4$
$\Rightarrow ∠ A = ∠ D$
Thus, in $ΔABC$ and $ΔDEF$ , we have
$∠ A = ∠ D$
and,
$\frac{AB}{DE} = \frac{AC}{DF}$
So, by $SAS$ -criterion of similarity, we have
$ΔABC ~ ΔDEF$ .

Weight(in kg)	Number of students $ΔABC ~ ΔDEF$
$40 ‐ 44$	$4$
$44 ‐ 48$	$6$
$48 ‐ 52$	$10$
$52 ‐ 56$	$14$
$56 ‐ 60$	$10$
$60 ‐ 64$	$8$
$64 ‐ 68$	$6$
$68 ‐ 72$	$2$

If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median of another triangle, then the two triangles are similar.

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