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