Let P be the midpoint of AC
Now, let
∠PAM = x°
∠PMA = y°
∠MAB = r°
∠MBA = s°
Since ∠C = 180-68-68 = 46°, ∠PMB = 180-46 = 134°
That gives you the equations
x+r = 68
y+s = 134
r+s+68 = 180
x+y = 90
Solve those, and m∠CMA = y+44 and m∠MAB = r
In isosceles △ABC with a base
AB
, the perpendicular bisector to
AC
, intersects the other leg
BC
at point M. If m∠B=68°, find: m∠CMA, m∠MAB
b
m∠MAB =
1 answer