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Ask QuestionPosted by Himanshu Parihar 6 years, 3 months ago
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Posted by Aman Aryan 6 years, 3 months ago
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Gaurav Seth 6 years, 3 months ago
Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that AB/PQ = AC/PR = AD/PM
To Prove: ΔABC ~ ΔPQR
Construction: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.
Proof: In ΔABD and ΔCDE, we have
AD = DE [By Construction]
BD = DC [∴ AP is the median]
and, ∠ADB = ∠CDE [Vertically opp. angles]
∴ ΔABD ≅ ΔCDE [By SAS criterion of congruence]
⇒ AB = CE [CPCT] ...(i)
Also, in ΔPQM and ΔMNR, we have
PM = MN [By Construction]
QM = MR [∴ PM is the median]
and, ∠PMQ = ∠NMR [Vertically opposite angles]
∴ ΔPQM = ΔMNR [By SAS criterion of congruence]
⇒ PQ = RN [CPCT] ...(ii)
Now, AB/PQ = AC/PR = AD/PM
⇒ CE/RN = AC/PR = AD/PM ...[From (i) and(ii)]
⇒ CE/RN = AC/PR = 2AD/2PM
⇒ CE/RN = AC/PR = AE/PN [∴ 2AD = AE and 2PM = PN]
∴ ΔACE ~ ΔPRN [By SSS similarity criterion]
Therefore, ∠2 = ∠4
Similarly, ∠1 = ∠3
∴ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P ...(iii)
Now, In ΔABC and ΔPQR, we have
AB/PQ = AC/PR (Given)
∠A = ∠P [From (iii)]
∴ ΔABC ~ ΔPQR [By SAS similarity criterion]
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-94+(-56)-(+98)+1900+(-654783)+78
= -94 -56 - 98+1900 -654783 +78
= -248 -654783 +1900 +78
= - 655031 + 1978
= -653053
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