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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 Triangle ABC similar to Triangle PQR.

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Answers

Answered by mahababu29
4

Question:

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 Triangle ABC similar to Triangle PQR.

Answer:

(Image 1)

Given: △ABC & △PQR

AD is the median of △ABC

PM is the median of △PQR

AB/PQ  =AC/PR  =AD/PM →1

To prove: △ABC∼△PQR

Proof: Let us extend AD to point D such that that AD=DE and PM upto point L such that PM=ML

Join B to E, C to E, & Q to L and R to L

(Image 2)

We know that medians is the bisector of opposite side

Hence BD=DC & AD=DE *By construction)

Hence in quadrilateral ABEC, diagonals AE and BC bisect each other at point D

∴ABEC is a parallelogram

∴AC=BE & AB=EC (opposite sides of a parallelogram are equal) →2

Similarly we can prove that

PQLR is a parallelogram.

PR=QL,PQ=LR (opposite sides of a parallelogram are equal) →3

Given that

AB/PQ=AC/PR =AD/PM  (from 1)

⇒AB/PQ  =BE/QL =AD/PM  (from 2 and 3)

⇒AB/PQ =BE/QL =2AD/2PM

​ ⇒AB/PQ =BE/QL=AE/PL

​  (As AD=DE,AE=AD+DE=AD+AD=2AD & PM=ML,PL=PM+ML=PM+PM=2PM)

∴△ABE∼△PQL (By SSS similarity criteria)

We know that corresponding angles of similar triangles are equal

∴∠BAE=∠QPL→4

Similarly we can prove that

△AEC∼△PLR

We know that corresponding angles of similar triangles are equal

∠CAE=∠RPL→5

Adding 4 and 5, we get

∠BAE+∠CAE=∠QPL+∠RPL

⇒∠CAB=∠RPQ→6

In △ABC and △PQR

AB/PQ  =AC/PR  (from 1)

∠CAB=∠RPQ (from 6)

∴△ABC∼△PQR (By SAS similarity criteria)

Hence, proved.

Attachments:
Answered by Anonymous
21

Produce AD to E so that AD = DE. Join CE

Similarly, produce PM to N such that PM = MN , and join RN.

In ΔABD and ΔCDE

AD = DE [By Construction]

BD = DC [AD is the median]

∠ADB = ∠CDE [Vertically opposite angles]

Therefore, ΔABD ≅ ΔECD [By SAS criterion of congruence]

⇒ AB = CE [CPCT] ...(i)

Also, in ΔPQM and ΔMNR

PM = MN [By Construction]

QM = MR [PM is the median]

∠PMQ = ∠NMR [Vertically opposite angles]

Therefore, ΔPQM = ΔNRM [By SAS criterion of congruence]

⇒ PQ = RN [CPCT]...(ii)

Now,

AB / PQ = AC / PR = AD / PM [Given]

⇒ 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 ]

Therefore, ΔACE ~ ΔPRN  [By SSS similarity criterion]

Therefore, ∠CAE = ∠RPN

Similarly, ∠BAE = ∠QPN

Hence, ∠CAE + ∠BAE = ∠RPN + ∠QPN

⇒ ∠BAC = ∠QPR

⇒ ∠A = ∠P ....(iii)

Now, In ΔABC and ΔPQR

AB/PQ = AC/PR

∠A = ∠P [from (iii)]

Therefore, ΔABC ~ ΔPQR  [By SAS similarity criterion]

Hope it helps!

Brainliest please!

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