To Prove:
|
A(∆ ABC)
|
=
|
AB2
|
=
|
BC2
|
=
|
AC2
|
||
A(∆ PQR)
|
PQ2
|
QR2
|
PR2
|
∴
|
AB
|
=
|
BC
|
=
|
AC
|
[C.S.S.T.]--------------------eq.
(1)
|
|
PQ
|
QR
|
PR
|
∴
|
AM
|
=
|
MB
|
=
|
AB
|
--------------------eq.
(3)
|
|
PN
|
NQ
|
PQ
|
Now,
|
A(∆ ABC)
|
=
|
½ (Base)(Height)
|
||||
A(∆ PQR)
|
½ (Base)(Height)
|
||||||
∴
|
A(∆ ABC)
|
=
|
BC × AM
|
||||
A(∆ PQR)
|
QR × PN
|
||||||
∴
|
A(∆ ABC)
|
=
|
AB × AB
|
[From eq. (1) and
(3)]
|
|||
A(∆ PQR)
|
PQ × PQ
|