If two sides and a median bisecting one of these sides of a triangle are respectively
proportional to the two sides and the corresponding median of another triangle, then provethat the two triangles are similar.
Answers
Answer
Step by step explanation: SOLUTION:
Given: In ∆ ABC and ∆PQR ,AD and PM are their medians.
AB/PQ = AC/PR= AD/PM…….(1)
TO PROVE:
∆ABC~∆PQR
Construction;
Produce AD to E such that AD=DE & produce PM to N such that PM= MN.
join BE,CE,QN,RN
PROOF:
Quadrilaterals ABEC and PQNR are parallelograms because their diagonals bisect each other at D and M.
BE= AC & QN= PR
BE/AC=1 & QN/PR=1
BE/AC =QN/PR or BE/QN = AC/PR
BE/QN= AB/PQ [ From eq1]
or AB/PQ= BE/QN…….(2)
From eq 1
AB/PQ= AD/PM= 2AD/2PM= AE/PN
[SINCE DIAGONALS BISECT EACH OTHER]
AB/PQ= AE/PN…………..(3)
From equation 2 and 3
AB/PQ=BE/QN= AE/PN
∆ABE ~∆PQN
∠1= ∠2…………..(4)
[Since corresponding angles of two similar triangles are equal]
Similarly we can prove that
∆ACE ~∆PRN
∠3=∠4…………(5)
ON ADDING EQUATION 4 AND 5
∠1+∠2=∠3+∠4
∠BAC = ∠QPR
and AB/PQ= AC/PR [from equation 1]
∆ABC~∆PQR
[By SAS similarity criteria]
HOPE THIS WILL HELP YOU....
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Answer:
Let us extend AD to point D such that AD=DE and PM up to point L such that PM=ML
Join B to E,C to E,Q to L and R to L
We know that medians are the bisector of the opposite side
Hence BD=DC
Also,AD=DE (by construction)Hence in quadrilateral ABEC, diagonals AE and BC bisect each other at point D
∴ quadrilateral ABEC is a parallelogram.
∴AC=BE and AB=EC .......(1) since opposite sides of a parallelogram are equal.
We know that medians are the bisector of the opposite side
Hence QM=MR
Also, PM=ML (by construction)Hence in quadrilateral PQLR, diagonals PL and QR bisect each other at point M
∴ quadrilateral PQLR is a parallelogram.
∴PR=RL and PQ=QL ........(1)since opposite sides of a parallelogram are equal.
Given:
PQ
AB
=
PR
AC
=
PM
AD
⇒
PQ
AB
=
QL
BE
=
PM
AD
from (1) and (2)
⇒
PQ
AB
=
QL
BE
=
2PM
2AD
⇒
PQ
AB
=
QL
BE
=
PL
AE
as AD=DE
AE=AD+DE=AD+AD=2AD
PM=ML
PL=PM+ML=PM+PM=2PM
∴△ABE∼△PQL by SSS similarity criterion
∴△ABE∼△PQL and △AEC∼△PLR
We know that corresponding angles of similar triangles are equal
∴∠BAE=∠QPL ........(3)
and ∠CAE=∠RPL ........(4)
Adding (3) and (4) we get
∠BAE+∠CAE=∠QPL+∠RPL
∠CAB=∠RPQ ........(5)
In △ABC and △PQR
PQ
AB
=
PR
AC
∠CAB=∠RPQ
∴△ABC∼△PQR by SAS similarity criterion
Step-by-step explanation: