Question

another question. please help no. 11

Answer 1

hii
given,
in ∆ PQR
PD_|_ QR,
PQ=a , PR= b, QD=c, and DR= d in the right angled ∆PQD
PQ^2 =PD^2 +QD^2( Pythagorous theorems)
a^2 =PD +c^2
PD^2 =a ^2-c^2 (I)

In right angled ∆PDR,
PR^2 =PD^2 +DR^2 ( Pythagorous theorems)
b^2 =PD^2 +d^2
PD^2 =b^2 -D^2 ......(I)

from equation (1) &(2)
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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