Question

in the fig Ac || ED and BE || CD
find
1) DFE
2) BDC​

Answer 1

DFC= 101

BDC=42

Step-by-step explanation:

let DFE be x

DEF+DFE+FDE=180°

DEF=BCE Alternate angles

therefore DEF=47°

So,47+x+32=180

180-(47+32)=x

x=101

BDC=EBD

ABE+EBD+FBC=180

106+X+32=180

SOLVING THIS WE GET 42