If two sides and median bisecting one of these sides of a triangle are respectively proportional to the two sides and the corresponding median of another triangle then prove that the triangles are similar.
Class - 10th
Chapter - Triangles
Kindly don't give absurd answers !!
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please provide all the steps along with the construction.
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Answers
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]
EXPLANATION.
Two sides and median bisecting one of these sides of a triangle are proportional to the two sides and the corresponding median of another triangle.
As we know that,
Construction:
Draw same triangle opposite to that triangle, we get.
To prove :
⇒ ΔABC = ΔPQR.
As we know that,
⇒ AX = AD + DX.
⇒ 2AD = AD + DX.
⇒ PY = PM + MY.
⇒ 2PM = PM + MY.
⇒ AB = CX.
⇒ AC = BX.
⇒ PQ = RY.
⇒ PR = QY.
By using Thales theorem, we get.
⇒ AB/PQ = AC/PR = AD/PM. --------(1).
We can also write this equation as,
⇒ AB/PQ = BX/QY = AD/PM. -------(2).
From equation (1) & (2), we get.
⇒ AB/PQ = BX/QY = 2AD/2PM.
We can also write as,
⇒ AB/PQ = BX/QY = AX/PY.
⇒ ΔABX ≈ ΔPQY [ SSS Criteria ].
⇒ ΔACX ≈ ΔPRY [ SSS Criteria ].
⇒ ∠BAX = ∠QPY. ---------(3).
⇒ ∠CAX = ∠RPY. ---------(4).
Adding equation (3) & (4), we get.
⇒ ∠BAX + ∠CAX = ∠QPY + ∠RPY.
⇒ ∠CAB = ∠RPQ. ---------(5).
⇒ In ΔABC = ΔPQR.
⇒ AB/PQ = AC/PR.
HENCE PROVED.
