192 K. R. S. Sastry
comparison of triangles BEC with BFC shows that
CE > BF. (1)
Complete the parallelogram BF GE. Since EG = BF, ∠F GE = B
2 , F G =
BE = CF implying that ∠F GC = ∠FCG. But by assumption ∠F GE = B
2 >
∠FCE = C
2 . So ∠EGC < ∠ECG, and CE < GE = BF, contradicting (1).
E
F
A G
B C
Figure 1.
Likewise, the assumption AB > AC also leads to a contradiction. Hence,
AB = AC and -
ABC must be isosceles.
3. The Gergonne analogue
We provide two proofs of Theorem 2 below. The first proof equates the ex-
pressions for the two Gergonne cevians to establish the result. The second one is
modelled on the proof of the Steiner - Lehmus theorem in §2 above.
Theorem 2. If two Gergonne cevians of a triangle are equal, then the triangle is
isosceles.
s − b s − c
s − c
s − a
s − a
s − b
I
A
B D C
E
F
Figure 2.