# In Fig. 3,<

By Pythagoras theorem

In ∆ ABC

AB2 = AC2 + BC2 …(1)

In ∆ ACD

AC2 = AD2 + DC2 …(2)

In ∆ BCD

BC2 = BD2 + CD2 …(3)

Add eq (2) and eq (3)

AC2 + BC2 = AD2 + DC2 + BD2 + CD2

AB2 = AD2 + 2CD2 + BD2

AB2 - AD2 - BD2 = 2CD2

BD(AB + AD – BD) = 2CD2

OR

Construction:
Join P to Q and A to Q.

In Δ ACQ,

AQ2 = AC2 + CQ2

In Δ PCB,

BP2 = PC2 + BC2

In ΔACB,

AB2 = AC2 + BC2

In ΔPCQ,

PQ2 = PC2 + CQ2

LHS:

AQ2 + BP2 = (AC2 + CQ2) + (PC2 + BC2)

RHS:

AB2 + PQ2 = (AC2 + BC2) + (PC2 + CQ2)

=(AC2 + CQ2) + (PC2 + BC2)

LHS = RHS

Hence proved.

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