In a triangle ABC let E be the midpoint of the side AC and F the midpoint of the side BC. Furthermore let G be the foot of the altitude through C on the side AB (or its extension). Show that the triangle EFG is isosceles if and only if ABC is isosceles.

**Solution 1 by Steve Dinh, a.k.a. Vo Duc Dien**

Let CG intercept EF at I. We only solve the problem with one geometrical configuration. Other configurations can also be solved similarly.

First assume that the triangle EFG is isosceles and GE = GF, ∠GEF = ∠GFE.

Since E and F are the midpoints of AC and BC, respectively, EF || AB and ∠GEF = ∠AGE and ∠GFE = ∠BGF . Combining with CG ⊥ AB, we have ∠EGC = ∠FGC. The two triangles EGC and FGC are then congruent since they also share GC. It follows that EC = FC and AC = BC or triangle ABC is isosceles.

Now assume triangle ABC is isosceles, AC = BC and EC = FC, ∠BAC = ∠ABC. Since ∠AGC = ∠BGC = 90°, two triangles AGC and BGC are congruent and AG = BG which leads us to ∠ACG = ∠BCG. The two triangles EGC and FGC are then congruent since they also share GC. It follows that EG = FG and triangle EFG is also isosceles.

**Solution 2**

First of all, since EF is the midline in ΔABC, EF||AB and, therefore, CG⊥EF, or, more importantly, GI⊥EF.

Assume ΔEFG is isosceles. Then, besides being the altitude from G, GI is also the median so that EI = EF. In ΔEFC, CI is both the median and the altitude from C, implying that the triangle is isosceles: CE = CF. From here, AC = BC, and ΔABC is also isosceles.

Assume ΔABC is isosceles. Then CG is both the altitude and the median from C. Since triangles ACG and ECI are similar as are triangles BCG and FCI, EI = FI. In ΔEFG, GI is both the altitude and the median from G, implying that the triangle is isosceles.