Question

In the given figure,
ΔFEC ≅ ΔGBD and /_1=/_2.
Prove that ΔADE ~ ΔABC.​

Answer 1

Answer:

if ∆Fec=~ ∆GBD SO THAT

DB =EC AND angle DBG = angle ECF

so that

AB = AC (because of equal angles DBG and ECF)

therefore

1=DBG & 2=ECF

and third angle A is common so it's congurent triangles by AAA