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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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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.
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!
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