Question

in figure 3.63 seg ad ⊥ seg bc
seg ae is the bisector of angle cab and c-e-d
prove that angle dae = 1/2 (angle c --angle b)
if you give the correct answer
i will mark you as brainliest

Answer 1

Step-by-step explanation:

As AE bisects ∠A, then,

∠BAE=∠CAE

In ΔADC,

∠ADC+∠DAC+∠ACD=180°

90°+∠DAC+∠C=180 °

∠C=90°−∠DAC

In ΔADB,

∠ADB+∠DAB+∠ABD=180°

90°+∠DAB+∠B=180°

∠B=90° −∠DAB

∠C−∠B=∠DAB−∠DAC

∠C−∠B=(∠BAE+∠DAE)−(∠CAE−∠DAE)

∠C−∠B=∠BAE+∠DAE−∠BAE+∠DAE

∠C−∠B=2∠DAE

1/2 (∠C−∠B)=∠DAE