In ∆PQR, PD 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).


Given,

In ∆PQR,


PDQR,


PQ = a, PR = b, QD = c and DR = d


in right angled ∆PDQ,


PQ2 = PD2 + QD2 [by Pythagoras theorem]


= a2 = PD2 + c2


= PD2 = a2 – c2 …..(i)



In right angled ∆PDR,


PR2 = PD2 + DR2 [by Pythagoras theorem]


= b2 = PD2 + d2


= PD2 = b2 - d2 ……(i)


From Equation (i) and (ii),


A2 - c2 = b2 - d2


A2 - b2 = c2 - d2


= (a – b) (a + b) = (c – d) (c + d)


Hence proved.


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