142 MATHEMATICS
We need to prove that
ar (ABC) AB^22 BC CA^2
ar (PQR) PQ QR RP✁ ✁ ✁
✂☎ ✆ ✂☎ ✆ ✂☎ ✆ ✄
✝ ✞ ✝ ✞ ✝ ✞
For finding the areas of the two triangles, we draw altitudes AM and PN of the
triangles.
Now, ar (ABC) =
1
BC × AM
2
and ar (PQR) =
1
QR × PN
2
So,
ar (ABC)
ar (PQR) =(^1) BC × AM
2 BC × AM
(^1) QR × PN
QR × PN
2
✟
✠
✟
(1)
Now, in ✡ ABM and ✡ PQN,
☛ B =☛ Q (As ✡ ABC ~ ✡ PQR)and ☛ M =☛ N (Each is of 90°)
So, ✡ ABM ~✡ PQN (AA similarity criterion)
Therefore,
AM
PN
=
AB
PQ (2)
Also, ✡ ABC ~✡ PQR (Given)
So,
AB
PQ =
BC CA
QR RP
☞ (3)
Therefore,
ar (ABC)
ar (PQR) =AB AM
PQ PN
✌ [From (1) and (3)]=
AB AB
PQ PQ
✌ [From (2)]=
AB^2
PQ
✁
☎ ✆
✝ ✞
Now using (3), we get
ar (ABC)
ar (PQR) =AB^22 BC CA^2
PQ QR RP
✁ ✁ ✁
☎ ✆ ✂☎ ✆ ✂☎ ✆
✝ ✞ ✝ ✞ ✝ ✞
✍
Let us take an example to illustrate the use of this theorem.