Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.
Given: ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR
i.e., 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]
∴ ∠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 equation (i) and (ii), we get
ΔABC ~ ΔPQR [By SAS similarity criterion]
