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.
