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 $∆ABC \sim ∆PQR$.

Given:

Two triangles $ΔABC$ and $ΔPQR$ in which $AB, AC$ and median $AD$ of $ΔABC$ are proportional to sides $PQ, PR$ and median $PM$ of $ΔPQR$

$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$

To do:

We have to prove that $ΔABC \sim ΔPQR$

Solution:

We have,

$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$ (D is the mid-point of BC and M is the mid point of QR)

$ΔABD \sim ΔPQM$ [SSS similarity criterion]

Therefore,

$\angle ABD = \angle PQM$ [Corresponding angles of two similar triangles are equal]

$\angle ABC = \angle PQR$

In $ΔABC$ and $ΔPQR$

$\frac{AB}{PQ} = \frac{AC}{PR}$........(i)

$\angle A = \angle P$........(ii)

From above equation (i) and (ii), we get,

$ΔABC \sim ΔPQR$ [By SAS similarity criterion]

Hence proved.

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Updated on: 10-Oct-2022

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