In the adjoining figure, D and E are respectively the midpoints of sid

1.	In the adjoining figure, D and E are respectively the midpoints of sides AB and AC of △ ABC. If PQ \|\| BC and CDP and BEQ are straight lines then prove that ar (△ ABQ) = ar (△ ACP).
Answer» 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).

In the adjoining figure, D and E are respectively the midpoints of sides AB and AC of △ ABC. If PQ || BC and CDP and BEQ are straight lines then prove that ar (△ ABQ) = ar (△ ACP).

Answer»

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).