In the figure, AP || BQ || CR. Prove that $a r (△ A Q C) = a r (△ P B R)$

Open in App

Solution

In given figure $A P | | B Q | | C R$
Then $Δ A B Q$ and $Δ P B Q$ lie on same base BQ and between parallel line AP and BQ
$∴ a r e a Δ A B Q = a r e a Δ P B Q$ ..........................(1)
And $Δ C Q B$ and $Δ R Q B$ lie on same base BQ and between parallel line CR and BQ
$∴ a r e a Δ C Q B = a r e a Δ R Q B$ ..........................(2)
Add (1) and (2) we get
$∴ a r e a Δ A B Q + a r e a Δ C Q B = a r e a Δ P B Q + a r e a Δ R Q B$
Hence $a r e a Δ A Q C = a r e a Δ P B R$

Suggest Corrections

Similar questions

A P ∥ B Q ∥ C R

a r (A Q C) = a r (P B R)

In the given figure, AP $∥$ BQ $∥$ CR. Prove that ar ( $△$ AQC) = ar ( $△$ PBR).
[2 MARKS]

∥

∥

△

In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

Join BYJU'S Learning Program