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Converse of Theorem 2

In the given ...

Question

In the given figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P. Then:

A r e a (Δ A B P) = A r e a (q u a d A B C D)

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A r e a (Δ A B C) = A r e a (Δ A C P)

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A r e a (Δ B C D) = A r e a (q u a d A C P D)

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Solution

The correct option is A $A r e a (Δ A B P) = A r e a (q u a d A B C D)$
Here $Δ D A C a n d Δ P A C$ are on the same base AC and between parallel lines AC and DP
$\Rightarrow A r e a (Δ D A C) = A r e a (Δ P A C)$

Adding $A r e a (Δ A C B)$ on both sides, we have,
$\Rightarrow A r e a (Δ D A C) + A r e a (Δ A C B)$
$= A r e a (Δ P A C) + A r e a (Δ A C B)$

$\Rightarrow A r e a (q u a d A B C D) = A r e a (Δ A P B)$

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Similar questions

A B C D

A

D

A C

B C

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(△ A B P) = a r e a (q u a d r i l a t e r a l A B C D)

A B C D

D

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(△ A B E) =

A B C D

□

△

△

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a r (Δ A B E) = a r q u a d . (A B C D)

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