Question

Bisectors of vertex angles A, B, and C of a triangle ABC intersect its circumcircle at the points D, E and F respectively. Prove that angle EDF  = 90° – `1/2` ∠A.

Answer 1

We have,

Given that, BE is the bisector of B.

`therefore angle ABE = (angleB)/2`

`angleADE = angleABE`   ...(Angles in the same segment for chord AF)

Similarly, `angle ACF = angleADF = (angleC)/2` ...(Angle in the same segment for chord AF)

`angleD = angleADE + angleADF`

`angleD = (angleB)/2 + (angleC)/2`

= `1/2  (angleB + angleC)`

= `1/2 (180^circ - angleA)`   ...`(∴ angleA + angleB + angleC = 180^circ)`

`angleD = 90^circ - (angleA)/2`