Isosceles Triangle Proof

[Clifford]

Triangle JKL is an isosceles triangle in which M is the midpoint of KL. Prove that M lies on the bisector of <J.

Would this proof be correct?

<J + <K + <M = 180 degrees
by drawing a median we get AM
(1)<J1 + <K + <M1 = 180 degrees
(2)<J2 + <L + <M2 = 180 degrees
<J1 = <J2 def'n of median
<K = <L def'n of an isosceles triangle
(1)=(2)
<J1 + <K + <M1 = <J2 + <L + <M2
subt in for J1 and K
<J2 + <L + <M1 = <J2 + <L + <M2
<M1 = <M2
<M1 = 90 supplementry angle theorem
<M2 = 90 supplementry angle theorem

Since <M1 and <M2 = 90 degrees, M lies on the bisector of <J

 

[pka]

\(\displaystyle \overline {JK} \equiv \overline {JL} ,\quad \overline {MK} \equiv \overline {ML} ,\quad \overline {MJ} \equiv \overline {MJ} \quad \Rightarrow \quad \Delta KMJ \equiv \Delta LMJ\)