The second question is beyond my understanding, and this is the answer I found online:
Take point h on AC so that AH=MN.
AM divides ∠BAC equally, so MH=MN
BM+MN=BM+MH
In order to minimize BM+MN=BM+MH, BM+MN=BM+MH=BH if and only if the straight lines H, M, B*** and BH are perpendicular to AC.
At this time △ABH is an isosceles right triangle.
BH= (radical number 2)/2 AB= (radical number 2)/2 * 4 Radical number 2=4
or
Take m as the fixed point first, and then when MN is the smallest, MN⊥AB.
Jean ME⊥AC
∵AD is the bisector of∝∠ ∝∠BAC.
∴ME=MN
In △BME, if BM+ME is the smallest, then B, M and E are on the same straight line.
∴BE⊥AC
∫∠BAC = 45 AB = 4∠2。
∴BE=4, that is, the minimum value of BM+MN is 4.
Because it is.