A
C
B
Ex.6.5 (Q.13)
Q. D & E are points on the sides CA & CB respectively
of ΔABC right angled at C.
Prove that : AE2 + BD2 = AB2 + DE2
ΔABC right angled at C
D & E are points on the sides CA & CB respectively
D
E
Proof :
In Δ ACE,
AE2 = AC2 + CE2
∠ ACE = 900
... (i) ... [by Pythagoras
theorem]
In Δ DCB,
∠ DCB = 900
BD2 = BC2 + CD2
... (ii) ... [by Pythagoras
theorem]
AE2 + BD2 = AC2 + CE2 + BC2 + CD2
... (iii) ... [by Adding (i)
and (ii)]
In Δ ACB,
AB2 = AC2 + BC2
∠ ACB = 900
... (iv) ... [by Pythagoras
theorem]
In Δ DCE,
∠ DCE = 900
DE2 = DC2 + CE2
... (v) ... [by Pythagoras
theorem]
AE2 + BD2 = AB2 + DE2
... [From (iii), (iv) & (v)]
AE2 + BD2 = AC2 + CE2 + BC2 + CD2 ... (iii)
AB2 = AC2 + BC2 ... (iv)