# In triangle ABC, the bisectors of <B and <C meet AC and AB at M and N respectively.?

If segments BM and CN have the same length, prove that triangle ABC is isosceles.

### 1 Answer

- Anonymous1 decade agoFavorite Answer
I call : AB = c, BC = a , CA = b

AN = dA, BM = dB,

AM=b1, MC=b2

BN=a1, NC=a2

Now two theorems of plane geometry say:

1) d²B = ac - b1*b2

d²A = bc - a1.a2

2) c : b = a1 : a2 or a1.c=a2.b

a : c = b2 : b1 or b2.c = b1.a

Given is:

d²(A) = b.c - a1.a2 = d²(B) =c.a - b1.b2

(b1+b2).c - a1.a2 = c(a1+a2) - b1.b2

b1.c + b2.c -a1.a2 = c.a1 + c.a2 - b1.b2

Now with b2.c = a.b1 and c.a2 = a1.b we have:

b1.c + a.b1 -a1.a2 = c.a1 + a1.b -b1.b2

c.b1 + a1.b1 + a2.b1 - a1.a2 = c.a1 + a1.b1 + a1.b2 - b1b2

c.b1 + a1.b1 + a2.b1 + b1.b2 = c.a1 + a1.b1 + a1.b2 + a1.a2

c.b1 +a2.b1 + b1.b2 = c.a1 +a1.b2 + a1.a2

b1(c+a2+b2) = a1(c+b2+a2)

So a1=b1 and ABM and ABN are congruent : <A = <B

Previewing what I did, I saw that I should have the lettering of the triangle wrong, sorry for that. (HvD)