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


Sides AB and AC and Triangles CBSE 10th important questions

Given that, In Δ ABC and Δ PQR,

\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}

Since AD is median of BC, hence BC = 2BD

Similarly, PM is median of QR, hence QR = 2QM

\therefore  \frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM}

or  \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}

\therefore  Δ ABD ~ Δ PQM

Hence, ∠ B = ∠ Q ……………. (i)

Now In Δ ABC and Δ PQR, we know that,

or  \frac{AB}{PQ} = \frac{BC}{QR}  (given)

∠ B = ∠ Q        from equation (i)

\therefore  Δ ABC ~ Δ PQR……….. Hence proved

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