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Ii DeltaPQR, PDbotQR such that D lies on...

Ii `DeltaPQR`, PD`bot`QR such that D lies on QR, if PQ=a,PR=b,QD=c and DR=d, then prove that (a+b)(a-b)=(c+d)(c-d).

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Given: In `trianglePQR, PD bot QR, so angle1= angle2`
PQ=a,PR=b,QD=c and DR=d
To prove: (a+b) (a-b) = (c+d) (c-d)
Proof : in right angle `trianglePDQ`
`PD^(2)= PQ^(2)-QD^(2)` (by pythagoras theorem)
` Rightarrow " " PD^(2)=a^(2) -c^(2)` ...(1)
similarlym in right angled `trianglePDR`
`PD^(2)= PR^(2)-DR^(2)` ( by pythagoras theorem) (by pythagoras theorem )
`PD^(2) = b^(2)-d ^(2)`
From (1) and (2) ,we have
`a^(2)-c^(2)= b^(2)-d^(2)`
`a^(2)-b^(2) = c^(2) -d^(2)`
(a-b) (a+b) = (c-d) (c+d) Hence proved.
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