Question

In the figure, AD and CE are bisectors of angle A and angle C respectively. If angle ABC= 90 degree, find angle ADC+ angle AEC.

Answer 1

We have ,

∠ABC + ∠BCA + ∠CAB = 180°

=>90° + ∠BCA + ∠CAB = 180°

=>∠BCA + ∠CAB = 180°

=> $\frac{1}{2}$ (∠BCA + ∠CAB ) = $\frac{1}{2}$ X 90°

=> $\frac{1}{2}$ (∠BCA + $\frac{1}{2}$ ∠CAB = 45°

=> ∠OCA + ∠OCA = 45°

=> ∠AOC + ∠OCA + ∠OCA = ∠AOC + 45° (Adding ∠AOC on both side )

=> 180° = ∠AOC + 45°

So, ∠AOC = 135° Ans ...

Please mark it as brainiest answer ........

Answer 2

in triangle ABC, angle ABC = 90
so, angle BAC + angle BCA = 90 (angle sum property)

Now, AD is the bisector of angle BAC, so angle OAC = 1/2 angle BAC
and CE is the bisector of angle BCA, so angle OCA = 1/2 angle BCA

In triangle AOC,
angle AOC + angle OAC + angle OCA = 180
or angle AOC + 1/2 angle BAC + 1/2 angle BCA = 180
or angle AOC + 1/2 (angle BAC + angle BCA) = 180
or angle AOC + 1/2 * 90 = 180
or angle AOC + 45 = 180
or angle AOC = 180 - 45 = 13