Question

in the figure CD is the angle bisector of ECB, B =ACE prove that ADC=ACD

Answer 1

Therefore we can see that angle ACD =ADC=x+y...hence proved

Answer 2

Answer:

Step-by-step explanation:

to prove:  ∠adc= ∠ acd

given:   ∠1= ∠2, ∠b= ∠3

in triangle ebc ,

∠6+ ∠b+ ∠1+ ∠2=180° {angle sum property}

replacing ∠ B with ∠ 3

∠6+ ∠1+ ∠2+ ∠3=180°

since ∠,1= ∠2 hence  ∠ 2+ ∠1 can be written as 2∠1

∠6+2 ∠1+ ∠3=180°------------{1}

now in triangle edc ,

∠4+ ∠6+ ∠1=180°{angle sum property}

∠6=180-{ ∠1+ ∠4}------------{2}

from{2}

180-{ ∠1+ ∠4}+2 ∠1+ ∠3=180°

2 ∠1+ ∠3= ∠1+ ∠4 + 180-180

2 ∠1- ∠1+ ∠3= ∠4

∠1+ ∠3= ∠4

hence proved , ∠ adc= ∠acd