If ΔABC and ΔDEF are triangles, ∠ABC ≅ ∠DEF, ∠BCA ≅ ∠EFD, segments AB ≅ DE, then ΔABC ≅ ΔDEF.

Question

If ΔABC and ΔDEF are triangles, ∠ABC ≅ ∠DEF, ∠BCA ≅ ∠EFD, segments AB ≅ DE, then ΔABC ≅ ΔDEF.

I was given the hint to use contradiction.

Assume ∠CAB not congruent to ∠FDE.

Without loss of generality, μ(∠CAB) < μ(∠FDE)

There is a point G on same side of line DE as F so that μ(∠EDG)=μ(∠CAB), by protractor postulate (angle construction)

By betweenness theorem for rays, ray DG is between ray DE and ray DF.

so ray DG meets segment EF at say H, by crossbar theorem.

Now from here i kind of get lost..

Answer 1

just observe that triangles are similar and one pair of lines is equal hence triangles are ongruent