Question 14 Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ∼ ΔPQR..............How many times has this question come for board exam ? Plz respond fast ....... thank you
Answers
Question 14 Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR.
Given,
In triangle ABC,
Sides AB, AC and Media AD
In triangle PQR,
Sides PQ, PR and Media PM
AB/PQ = AD/PM = AC/PR
To prove: ΔABC ∼ ΔPQR
Construction: Produce AD to E such that, AD = DE
Produce PM to N such that, PM = MN
Proof:
In Δ ABD and Δ EDC
AD = DE (construction)
∠ ADB = ∠ EDC (vertically opposite angles)
BC = DC (AD is a median)
∴ Δ ABD ≅ Δ EDC (by SAS theorem)
∴ AB = CE (c.p.c.t)
Similarly, PQ = RN and ∠ 1 = ∠2
AB/PQ = AD/PM = AC/PR (given)
⇒ CE/RN = 2AD/2PM = AC/PR
⇒ CE/RN = AE/PN = AC/PR
∴ Δ AEC ~ Δ PNR (by SSS theorem)
⇒ ∠ 3 = ∠ 4 (∵ ∠ 1 = ∠ 2 )
∴ ∠ 1 + ∠ 2 = ∠ 3 + ∠ 4
∴ ΔABC ∼ ΔPQR (by SAS theorem)
Hence the proof.
