In the adjoining figure, $D$ and $E$ are respectively the midpoints of sides $A B$ and $A C$ of $△ A B C$ . If $P Q ∥ B C$ and $C D P$ and $B E Q$ are straight lines, then prove that $a r (△ A B Q) = a r (△ A C P)$ .

Open in App

Solution

In $△ P A C$
$P A ∥ D E$ and $E$ is the midpoint of $A C$
So, $D$ is the midpoint of $P C$
By Converse of midpoint theorem
Also, $D E = \frac{1}{2} P A$ ----- $(1)$
Similarly, $D E = \frac{1}{2} A Q$ ------- $(2)$
From $(1)$ and $(2)$ we have $P A = A Q$
$△ A B Q$ & $△ A C P$ are on same base PQ and between same parallels PQ and BC
$∴ a r (△ A B Q) = a r (△ A C P)$ (proved)

Suggest Corrections

Similar questions

In the adjoining figure,D and E are respectively the midpoints of sides AB and AC of $△ A B C$ .If PQ||BC and CDP and BEQ are straight lines then prove that $a r (△ A B Q) = a r (△ A C P)$ .

Q. 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).

Q. In the adjoining figure, D, E, F are the midpoints of the sides BC, CA and AB respectively, of ∆ABC. Show that ∠EDF = ∠A, ∠DEF = ∠B and ∠DEF = ∠C.
Figure

Q. In the adjoining figure,

D, E, F

are the midpoints of the sides

B C, C A a n d A B

Δ A B C

. If

B E a n d D F

intersect at

X

while

C F a n d D E

intersect at

Y

, prove that

X Y = \frac{1}{4} B C

In the adjoining figure, D, E, F are the midpoints of the sides BC, CA and AB respectively, of $Δ A B C$ . Show that $∠ E D F = ∠ A, ∠ D E F = ∠ B a n d ∠ D F E = ∠ C$ .

Join BYJU'S Learning Program

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).

In △PACPA∥DE and E is the midpoint of ACSo, D is the midpoint of PC By Converse of midpoint theoremAlso, DE=12PA ----- (1)Similarly, DE=12AQ -------(2)From (1) and (2) we have PA=AQ△ABQ & △ACP are on same base PQ and between same parallels PQ and BC∴ar(△ABQ)=ar(△ACP) (proved)

In the adjoining figure, $D$ and $E$ are respectively the midpoints of sides $A B$ and $A C$ of $△ A B C$ . If $P Q ∥ B C$ and $C D P$ and $B E Q$ are straight lines, then prove that $a r (△ A B Q) = a r (△ A C P)$ .