D. Given: in a A PQR. (PQR= 90°and seg. QS_1 seg. PR, where S lies on PR.
Prove that PQP+QR2 =PR?
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Answers
Answer:
Given: ΔPQR is right angled triangle where Q is right angle
QS ⊥ PR
QM bisect the ∠PQR
To prove: \frac{PM^2}{MR^2}=\frac{PS}{SR}
MR
2
PM
2
=
SR
PS
In ΔPQR,
QM bisector of ∠PQR
\implies\frac{PM}{MR}=\frac{PQ}{QR}⟹
MR
PM
=
QR
PQ
(Property of angle bisector of a triangle)
\implies\frac{PM^2}{MR^2}=\frac{PQ^2}{QR^2}⟹
MR
2
PM
2
=
QR
2
PQ
2
..............(1)
Again in ΔPQR
QS ⊥ PR ⇒ ΔPQR is similar to ΔPSQ is similar to ΔRSQ
in ΔPSQ and ΔPQR
[tex\frac{PQ}{PR}=\frac{PS}{PQ}[/tex] (both triangles are similar)
PQ^2=PR\times PSPQ
2
=PR×PS .........(2)
Similarly form ΔPSQ and ΔPQR
QR^2=PR\times SRQR
2
=PR×SR ..............(3)
From (1) , (2) & (3)
\frac{PM^2}{MR^2}=\frac{PR\times PS}{PR\times SR}
MR
2
PM
2
=
PR×SR
PR×PS
\frac{PM^2}{MR^2}=\frac{PS}{SR}
MR
2
PM
2
=
SR
PS