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).

Explanation

Given In `DeltaPQR,PDbotQR,PQ`=a,PR=b,QD=c and DR=d
To prove (a+b)(a-b)=(c+d)(c-d)
Proof In right angled `DeltaPDQ`
`PQ^(2)=PD^(2)+QD^(2)` [by pythagoras theorem]
`rArr a^(2)=PD^(2)+c^(2)`
`rArr PD^(2)=a^(2)-c^(2)`

In right angled `DeltaPDR PR^(2)=PD^(2)+DR^(2)` [by Pythagoras theorem]
`rArr b^(2)=PD^(2)+d^(2)`
`rArr PD^(2)=b^(2)-d^(2)`
From Eqs, (i) and (ii),
`a^(2)-c^(2)=b^(2)-d^(2)`
`rArr a^(2)-b^(2)=c^(2)-d^(2)`
`rArr (a-b)(a+b)=(c-d)(c+d)` Hence proved