Question

Sides AB and BC and median AD of a
triangle ABC are respectively propor-
tional to sides PO and OR and median
PM of triangle PQR (see Fig. 6.41). Show that triangle ABC is similar to triangle PQR​

Answer 1

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]

Answer 2

Answer:

$<font color="red">$

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

$<font color="blue">$

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]