If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median of another triangle then prove that the two Triangles are similar.
Or
Side AB and AC and median AD of triangle ABC are respectively proportional to sides PQ and PR and Median PM of another triangle PQR show that ∆ABC~∆PQR.
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)
Answers
Answered by
100
[FIGURE IS IN THE ATTACHMENT]
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....
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....
Attachments:
vsupriya070402:
ur wish if u want easy answer refer to my answer if u want all the construction means take this answer
ur answer is wrong @vsupriya070402
but how can u say that
I think ur in 10th
who is in 10th??
what ur studying @nikitasingh79
I m a teacher not a student...
OK
for which class ur teaching
Answered by
30
ANSWER
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BD = DC and QM = MR
Also, AD = DE (By construction)
And, PM = ML (By construction)
In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.
Therefore, quadrilateral ABEC is a parallelogram.
∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)
Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR
It was given that
∴ ΔABE ∼ ΔPQL (By SSS similarity criterion)
We know that corresponding angles of similar triangles are equal.
∴ BAE = QPL (1)
Similarly, it can be proved that ΔAEC ∼ ΔPLR and
CAE = RPL (2)
Adding equation (1) and (2), we obtain
BAE + CAE = QPL + RPL
⇒ CAB = RPQ (3)
In ΔABC and ΔPQR,
CAB = RPQ [Using equation (3)]
∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)
..........
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BD = DC and QM = MR
Also, AD = DE (By construction)
And, PM = ML (By construction)
In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.
Therefore, quadrilateral ABEC is a parallelogram.
∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)
Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR
It was given that
∴ ΔABE ∼ ΔPQL (By SSS similarity criterion)
We know that corresponding angles of similar triangles are equal.
∴ BAE = QPL (1)
Similarly, it can be proved that ΔAEC ∼ ΔPLR and
CAE = RPL (2)
Adding equation (1) and (2), we obtain
BAE + CAE = QPL + RPL
⇒ CAB = RPQ (3)
In ΔABC and ΔPQR,
CAB = RPQ [Using equation (3)]
∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)
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