Question

$\huge\underline\bold\red{Question}$
Sides AB and BC and Median AD of a triangle ABC are proportional to sides PQ AND PR and median PM of triangle PQR. Prove triangle AbC~trinangle PQR .
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Answer 1

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

AB/PQ = BC/QR = AD/PM

To Prove: ΔABC ~ ΔPQR

Proof: AB/PQ = BC/QR = AD/PM

AB/PQ = BC/QR = 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

AB/PQ = BC/QR ———(i)

∠ABC = ∠PQR ——-(ii)

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

ΔABC ~ ΔPQR [By SAS similarity criterion]

Hence Proved

Answer 2

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

AB/PQ = BC/QR = AD/PM

To Prove :

ΔABC ~ ΔPQR

Proof :

AB/PQ = BC/QR = AD/PM

AB/PQ = BC/QR = 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

AB/PQ = BC/QR ———(i)

∠ABC = ∠PQR ——-(ii)

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

ΔABC ~ ΔPQR [By SAS similarity criterion]

Hence Proved !!