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Question
In Δ ABC and Δ PQR,
∠ ABC ≅ ∠ PQR, seg BD and
seg QS are angle bisector.
`If (l(AD))/(l(PS)) = (l(DC))/(l(SR))`
Prove that : Δ ABC ∼ Δ PQR
Solution
Proof : `(l(AD))/(l(PS)) = (l(DC))/(l(SR))` ∴ `(l(AD))/(l(DC)) = (l(PS))/(l(SR))`
According to angle bisector theorem, `(l(AD))/(l(DC)) = (l(AB))/(l(BC)) ; (l(PS))/(l(SR)) = (l(PQ))/((QR))`
`∴ (l(AB))/(l(BC)) = (l(PQ))/(l(QR))` and ∠ ABC ≅ ∠ PQR ..... (Given)
Δ ABC ∼ Δ PQR ........ (SAS Test)
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∴ `square/square = square/square` .......... (II) theorem of angle bisector.
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∴ `("AE")/("EB") = ("AD")/("DC")` ....(ii) `square`
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In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR.
Complete the proof by filling in the boxes.
solution:
In ∆PMQ,
Ray MX is the bisector of ∠PMQ.
∴ `("MP")/("MQ") = square/square` .............(I) [Theorem of angle bisector]
Similarly, in ∆PMR, Ray MY is the bisector of ∠PMR.
∴ `("MP")/("MR") = square/square` .............(II) [Theorem of angle bisector]
But `("MP")/("MQ") = ("MP")/("MR")` .............(III) [As M is the midpoint of QR.]
Hence MQ = MR
∴ `("PX")/square = square/("YR")` .............[From (I), (II) and (III)]
∴ XY || QR .............[Converse of basic proportionality theorem]
