Triangle ADC and ABC are right triangles.

AC shares ∠MAN. ∠MAN= 120 degrees.

So the angles CAD and CAB = 60

So AD=0.5AC

AB = 0.5AC communication

So AB+AD=AC

2 established

And point c is perpendicular to AM and an, and intersects with p and q respectively.

Since ∠ABC+∠ADC= 180 and ∠DCB=60, this is fixed because the sum of the internal angles of the quadrilateral is 360.

∠PCQ=60

So PCD=∠QCB.

PD=CPtg∠PCD，QB=CQtg∠QCB

∠PCD=∠QCB, CP=CQ, triangle hat, CAQ congruence, so CP=CQ.

So PD=QB

AB+AD=PA-PD+AQ+QB=PA+AQ=AC (the first question has been proved).

You wrote the third question correctly, didn't you?

∠ABC+∠ADC= 180

If so.

According to the conclusion of the second question, AB+AD=AC direct.