Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(Δ APB) ✕ ar (Δ CPD) = ar (Δ APD) ✕ ar (Δ BPC)

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Solution

Given:

(1) ABCD is a quadrilateral,

(2) Diagonals AC and BD of quadrilateral ABCD intersect at P.

To prove: Area ofΔ APB ×Area of ΔCPD = Area of ΔAPD × Area of ΔBPC

Construction: Draw AL perpendicular to BD and CM perpendicular to BD

Proof:

We know that

Area of triangle = × base× height

Area of ΔAPD = × DP × AL …… (1)

Area of ΔBPC = × CM × BP …… (2)

Area of ΔAPB = × BP × AL …… (3)

Area of ΔCPD = × CM × DP …… (4)

Therefore

Hence it is proved that

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Similar questions

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:

ar $(△ A P B) \times a r (△ C P D) = a r (△ A P D) \times a r (△ B P C)$

a r (A P B) \times a r (C P D) = a r (A P D) \times a r (B P C)

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that

[Hint: From A and C, draw perpendiculars to BD]

a r (A P B) \times a r (C P D) = a r (A P D) \times a r (B P C)

a r (A P B) \times a r (C P D) = a r (A P D) \times a r (B P C)

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