Sides AB and BC and Median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\triangle$PQR . Show that $\triangle$ABC $\triangle$PQR
Given: Two triangles. ΔABC and ΔPQR in which AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$
To Prove: ΔABC ~ ΔPQR
Solution:
We have $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$ (D is the mid-point of BC. M is the mid point of QR)
ΔABD ~ ΔPQM [SSS similarity criterion]
Therefore, ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]
∠ABC = ∠PQR
In ΔABC and ΔPQR
$\frac{AB}{PQ} = \frac{BC}{QR}$ ———(i)
∠ABC = ∠PQR ——-(ii)
From above equation (i) and (ii), we get
ΔABC ~ ΔPQR [By SAS similarity criterion]
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