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In the adjoining figure, D,E and F are t...

In the adjoining figure, D,E and F are the mid- points of the sides of `DeltaPQR` and 'S' is the fooot of perpendicualar form P to side S.
Prove that:
(i) `FQ=FS` and `angleFQS=angle FSQ`.
(ii)` angle FQS=angleFED`.
(iii) square `FSDE` is a cyclic quadrilateral.

Text Solution

Verified by Experts

(i) `PS bot QR`
`rArrangle PSQ=90^@`
F is the mid-point of PQ.
`therefore FS =(1)/(2)PQ` or `FS=FP`
Now, `DeltaPQS` is right-angled triangle and we know that the mid-point of hypotenuse is equidistant form the vertices of a triangle.
`thereforeFQ=FP =FS`
`rArrFQ=FS`
`rArrangleFSQ=angleFQS` ..........(1)
(ii) F is the mid-point of PQ and E is the mid-point of PR.
`thereforeFE|\ |QR`
`rArrFE|\ |QD`
Similarly, `FQ|\ |ED`
`therefore square FEDQ` is a parallelogram.
We know that opposite angles of a parallelogram are equal.
`therefore angle FQS=angle FED`
(iii) Now from eqs. (1) and (2),
`angle FSQ=angleFED`
But `angleFSQ` is the exterior angle of square `FSDE`.
square ` FSDE` is a cyclic quadrilateral.
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