Subject :IMO Class : Class 10
In the given adjacent figure, BA and BC are produced to meet CD and AD produced in E and F. Then ∠AED + ∠CFD =
A 80°
B 50°
C 40°
D 160°
Ans 1:
Class : Class 10
C
Ans 2: (Master Answer)
Class : Class 1
The correct answer is C.
In ΔAED,
∠AE D + ∠EAD + 90° = 180° (Angle sum property)
⇒ ∠AED + ∠EAD = 90°
⇒ ∠EAD = 90° – ∠AED ...(i)
Also, ∠EAD = (50° + ∠AFB) ...(ii) (Exterior angle property)
From (i) and (ii), we have
50° + ∠AFB = 90° – ∠AED
⇒ ∠AFB + ∠AED = 90° – 50° = 40°
⇒ ∠AED + ∠CFD = 40°