Question

Sides ab,ac and median ad of a triangle are respectively proportinal to sides pq,pr and median pm of another triangle pqr. Show that tria

Answer 1

Median divides the opposite side.

∴

Given that,

In ΔABD and ΔPQM,

(Proved above)

∴ ΔABD ∼ ΔPQM (By SSS similarity criterion)

⇒ ∠ABD = ∠PQM (Corresponding angles of similar triangles)

In ΔABC and ΔPQR,

∠ABD = ∠PQM (Proved above)

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

Answer 2

Answer:

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

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