the side AB,BC and median AD of the triangle ABC are respectively prop

1.	the side AB,BC and median AD of the triangle ABC are respectively proprotionl to PQ, QR and median of triangle PQR
Answer» 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] 👍👍thank you👍👍

the side AB,BC and median AD of the triangle ABC are respectively proprotionl to PQ, QR and median of triangle PQR

Answer»

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]

👍👍thank you👍👍