5.
We are given that AB ≅ CB and AD ≅ CD.
We know that BD ≅ BD, so ΔABD ≅ ΔCBD by SSS.
Then ∠1 ≅ ∠2 by CPCTC; and ∠3 ≅ ∠4, also by CPCTC.
Since ∠1 ≅ ∠2, BD is the bisector of (∠1+∠2) = ∠ABC.
Since ∠3 ≅ ∠4, BD is the bisector of (∠3+∠4) = ∠ADC.
6.
From (5), we know BD is the angle bisector of ∠ABC. ΔABC is isosceles, so the bisector of the apex angle is also an altitude and median. That is, point M on BD and AC must be the midpoint of AC by the definition of a median.
