Let Delta ABC be a triangle and Delta DEF be a triangle Let

Let Delta ABC be a triangle and Delta DEF be a triangle. Let G be a point on the intersection of and ED. Let 1, 2, 3, and 4 be labeled as in the figure. Also, let 2 3, BG EG, and AD CF. Prove AB EF. Given: Prove: Proof:

Solution

In triangle GDC : <2 congruent to <3 => GC congruent to DG (Opposite angles of a triangle are congruent then opposite sides are equal, they share a common base)
and it is given BG congruent to EG. Therefore we can say, (BG + GC) congruent to (EG+GD) as GC = GD. Hence BC congruent to DE.
Also given AD congruent to CF. We can say (AD+DC) congruent to (CF+DC). Hence AC congruent to DF
Now in triangle ABC and DEF
<2 congruent to <3 , BC congruent to DE , AC congruent to DF
We can say now, triangle ABC and DEF are congruent. (SAS property)
Hence AB congruent to EF.