From the figure we know that D and E are the midpoints of AB and AC
So we get
DE || BC || PQ
Consider △ ACP
We know that AP || DE and E is the midpoint of AC
Using the midpoint theorem we know that D is the midpoint of PC
So we get
DE = ½ AP
It can be written as
AP = 2DE …… (1)
Consider △ ABQ
We know that AQ || DE and D is the midpoint of AB
Using the midpoint theorem we know that E is the midpoint of BQ
So we get
DE = ½ AQ
It can be written as
AQ = 2DE ……. (2)
Using equations (1) and (2)
We get
AP = AQ
We know that △ ACP and △ ABQ lie on the bases AP and AQ between the same parallels BC and PQ
So we get
Area of △ ACP = Area of △ ABQ
Therefore, it is proved that ar (△ ABQ) = ar (△ ACP).