Math, asked by swastikpodar15, 8 months ago

In triangle PQR,PM is the median , show that PQ+ QR+PR > 2PM

Answers

Answered by Mrnil07devil
0

Answer:

I don't know

Step-by-step explanation:

any one help him

Answered by parikshit17
3

Answer:

Given that triangle ABC and ΔPQR in which AD and PM are medians drawn on sides BC and QR respectively.

It is given that AB/PQ = AC/PR = AD/PM

We have to prove that ΔABC ~ ΔPQR

Construction: Produce AD to E such that AD = DE and PM to F such that PM = MF

From the figure,

In ΔABD and ΔCDE,

AD = DE [by construction]

∠ADB = ∠CDE [vertically opposite angles]

BD = DC [Since AD is a median]

So, by SAS congruent condition

ΔABC ≅ ΔPQR

AB = CE [by CPCT]

Similarly, we can prove

ΔPQM ≅ ΔRMF

PQ = RF [by CPCT]

Now, given that

AB/PQ = AC/PR = AD/PM

CE/RF = AC/PR = 2AD/2PM

CE/RF = AC/PR = AE/PF [Since AE = AD + DE and AD = DE, Same for PF]

By using SSS Congruent condition

ΔACE ≅ ΔPRF

=> ∠1 = ∠2 ......1

Similarly, ∠3 = ∠4 ......2

Adding equations 1 and 2, we get

∠1 + ∠3 = ∠2 + ∠4

=> ∠A = ∠P

Now, in ΔABC and ΔPQR

AB/PQ = AC/PR

and ∠A = ∠P

By SAS similar condtion,

ΔABC ~ ΔPQR

Similar questions