Question

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.
Sow that $\Delta ABC\sim \Delta PQR$.

Answer 1

(Image 1)
Given: $△ABC$ & $△PQR$
AD is the median of $△ABC$
PM is the median of $△PQR$
$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}\to 1$
To prove: $△ABC\sim △PQR$
Proof:Let us extend AD to point D such that that $AD=DE$ and PM upto point L such that $PM=ML$
Join B to E, C to E, & Q to L and R to L
(Image 2)
We know that medians is the bisector of opposite side
Hence $BD=DC$ & $AD=DE$ *By construction)
Hence in quadrilateral ABEC, diagonals AE and BC bisect each other at point D
$\therefore ABEC$ is a parallelogram
$\therefore AC=BE$ & $AB=EC$ (opposite sides of a parallelogram are equal) $\to 2$
Similarly we can prove that
$PQLR$ is a parallelogram.
$PR=QL,PQ=LR$ (opposite sides of a parallelogram are equal) $\to 3$
Given that
$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$ (frim $1$)
$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{AD}{PM}$ (from $2$ and $3$)
$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{2AD}{2PM}$
$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{AE}{PL}$ (As $AD=DE,AE=AD+DE=AD+AD=2AD$ & $PM=ML,PL=PM+ML=PM+PM=2PM$)
$\therefore △ABE\sim △PQL$ (By SSS similarity criteria)
We know that corresponding angles of similar triangles are equal
$\therefore \angle BAE=\angle QPL\to 4$
Similarly we can prove that
$△AEC\sim △PLR$
We know that corresponding angles of similar triangles are equal
$\angle CAE=\angle RPL\to 5$
Adding $4$ and $5$, we get
$\angle BAE+\angle CAE=\angle QPL+\angle RPL$
$\Rightarrow \angle CAB=\angle RPQ\to 6$
In $△ABC$ and $△PQR$
$\frac{AB}{PQ}=\frac{AC}{PR}$ (from $1$)
$\angle CAB=\angle RPQ$ (from $6$)
$\therefore △ABC\sim △PQR$ (By SAS similarity criteria)
Hence, proved