Question

Question 14
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that $\Delta ABC\sim \Delta PQR$.

Answer 1

Given, two triangles $\Delta ABC$ and $\Delta PQR$ in which AD and PM are medians such that $\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$
To Prove that $\Delta ABC\sim \Delta PQR$
Construction: Produce AD to E so that AD = DE. Join CE.
Similarly produce PM to N such that PM = MN, also Join RN.
Proof
In $\Delta ABD$ and $\Delta CDE$, we have
AD = DE [By Construction]
BD = DC [$\because$ AD is the median]
And, $\angle ADB=\angle CDE$ [Vertically opposite angles]
$\therefore \Delta ABD\cong \Delta CED$ [By SAS criterion of congruence]
$\Rightarrow AB=CE\left[byCPCT\right]...\left(i\right)$
Also, in $\Delta PQMand\Delta MNR$, we have
PM = MN [By Construction]
QM = MR [$\therefore PM$ is the median]
And, $\angle PMQ=\angle NMR$ [Vertically opposite angles]
$\therefore \Delta PQM=\Delta MNR$ [By SAS criterion of congruence]
$\Rightarrow PQ=RN\left[CPCT\right]...\left(ii\right)$
Now,$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$
$\Rightarrow \frac{CE}{RN}=\frac{AC}{PR}=\frac{AD}{PM}...\left[From\right(i\left)and\right(ii\left)\right]$
$\Rightarrow \frac{CE}{RN}=\frac{AC}{PR}=\frac{2AD}{2PM}$
$\Rightarrow \frac{CE}{RN}=\frac{AC}{PR}=\frac{AE}{PN}[\therefore 2AD=AEand2PM=PN]$
$\therefore \Delta ACE\sim \Delta PRN$ [By SSS similarity criterion]
Therefore, $\angle 2=\angle 4$
Similarly, $\angle 1=\angle 3$
$\therefore \angle 1+\angle 2=\angle 3+\angle 4$
$\Rightarrow \angle A=\angle P...\left(iii\right)$
Now, in $\Delta ABCand\Delta PQR,wehave$
$\frac{AB}{PQ}=\frac{AC}{PR}\left(Given\right)$
$\angle A=\angle P\left[From\right(iii\left)\right]$
$\therefore \Delta ABC\sim \Delta PQR$ [By SAS similarity criterion]