Math, asked by puja123roy456, 11 months ago

side ab and AC and median ad of a triangle ABC are respectively proportional to sides PQ and PR and medium PM of another triangle pqr so that triangle ABC similar Triangle pqr ​

Answers

Answered by basavaraj5392
2

Step-by-step explanation:

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

Answered by psupriya789
0

Answer:

Given two triangles. ΔABC and ΔPQR in which AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR

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]

Therefore, ∠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 above equation (i) and (ii), we get

ΔABC ~ ΔPQR [By SAS similarity criterion]

Hence Proved

HOPE IT HELPS U

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