Make a straight line PQ, cross E, and make ET perpendicular to BA.

E makes EH perpendicular to CN and e makes EK perpendicular to MN, because

EM bisects ∠BMN and EN bisects the angle MNC, so TE=KE=HE.

When the included angle APQ between PQ and AB is acute, TM=MK (congruence formula is available), because EM bisects ∠BMN.

Similarly, EN bisects the angle MNC, so KN=HN=HQ plus QN.

And because TE=KE=HE, it is easy to know that HQ=PT (congruent availability), so MN=MK+KN=TM+HN=TM+HQ+QN=

TM+PT+QN=PM+QN

So MN=PM+QN

Similarly, when the included angle APQ between PQ and AB is obtuse, MN=PM+QN can also be obtained.

When the intersection is discussed below,

Point E is PQ, and PQ and MN meet at point U.

Then make EJ perpendicular to BA through e.

E makes EL perpendicular to CN and e makes EO perpendicular to MN.

It is easy to know that JE=EO=EL.

Extend the EM intersection CD to Z, because JE=EO=EL, it is easy to know that ZL=JM (congruence is known).

Similarly, MEP and ZEQ are congruent, so QZ=MP, so NQ=NL+LZ+ZQ=NL+JM+MP.

Because EM bisects ∠BMN and EN bisects ∠MNC, it is known from congruence that LN=NO and JM=MO.

MN=MO+ON=JM+LN

NQ = Netherlands +LZ+ZQ = Netherlands +JM+MP.

So MN+MP=NQ