In the figure (1), add AP.
S△ABC=S△ABP+S△APC
1/2×BC×AM = 1/2×ab×DP+ 1/2×AC×PE
According to the meaning of the question, we can see that
AM=h,DP=h 1,PE=h2,PF=h3
∫h 1+H2+H3 = h,h3=0
∴h 1+h2=0
∴BC×h=AB×h 1+AC×h2
AB = AC
∴BC×h=AB×(h 1+h2)
BC×h=AB×h
BC=AB=AC
That is, △ABC is an equilateral triangle.
In figure (2),
S△ABC=S△ABP+S△APC+ S△PBC
1/2×BC×AM = 1/2×ab×BP+ 1/2×AC×PE+ 1/2×BC×PF
BC = AB = AC
∴AM=DP+PE+PF (equal property)
H=h 1+h2+h3 (equivalent substitution)
Similarly, in Figure (3),
h=h 1+h2+h3