Question

in the following figure, ∆FEC

Answer 1

Since ∆FEC is congruent to ∆GBD
THEREFORE. angleDBC = angleECB
and DB = EC

now DB = EC and AD=AE
therefore DB/EC = AD/AE
or AD/DB = AE/EC

therefore DE ll BC
therefore angle1 = angle3
angle2= angle4

so, ∆ADE ~ ∆ABC