Consider this blue triangle:
h$ so the area of the large blue and red right-angled triangle is half this, namely $\frac12(a+b)h$
The area of the small rectangle on the left is $ah$ so the area of the small red right-angled triangle on the left is half this, namely $\frac12ah$
So the area of the blue triangle is the difference in area between the two right-angled triangles, namely $\frac12(a+b)h-\frac12ah = \frac12bh$, i.e. half of base times height