Question

in the following figure , ab = bc , bc=cd and de is parallel to bc . calculate:
angle CDE
angle DCE

i want the answer of Q3

Answer 1

Answer:

CDE=52; DCE =12

Step-by-step explanation:

AB=AC=> ABC=ACB=(180-128)/2=64

BC=CD =< CDB =CBD = 64

DE is parralel BC so ADE=ABC=64 and AED=ACB=64

=> EDC=180-(64*2)=52

AED+DEC=180 => DEC=180-64=116

=> DCE=180-116-52=12

Answer 2

Answer:

128°+angle EAG=180°

EAG=52°

In triangle ABC,

AB=AC

Therefore,

Angle BAC =angle ACB

angle ACB=52°

In Triangle ABC,

A+B+C=180°(Angle Sum Property)

52°+52°+B=180°

B=180°-104°

B=76°

Now,

In triangle CDB,

CD=CB

therefore,

CBD=CDB=76°

By Angle Sum Property Of triangles,

DCB=180°-152°

DCB=28°

DE||BC,DC as transversal

DCB=CDE

Angle CDE=28°

We Know angle ACB=52°

Therefore

angle DCE+angle DCB=52°

angle DCE+28°=52°

angle DCE =52°-28°

Angle DCE=24°