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In the Following Figure, Ce is Drawn Parallel to Diagonals Db of the Quadrilateral Abcd Which Meets Ab Produced at Point E. Prove that δAde and Quadrilateral Abcd Are Equal in Area

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Question

In the following figure, CE is drawn parallel to diagonals DB of the quadrilateral ABCD which meets AB produced at point E.
Prove that ΔADE and quadrilateral ABCD are equal in area.

Sum
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Solution

Since ΔDCB and ΔDEB are on the same base DB and between the same parallels i.e. DB // CE, therefore we get

Ar. ( ΔDCB) = Ar. ( ΔDEB )
Ar. ( ΔDCB + ΔADB ) = AR. (ΔDEB + ΔADB )
Ar. ( ABCD ) = Ar. ( ΔADE )
Hence proved.

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Figures Between the Same Parallels
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Chapter 15: Area Theorems [Proof and Use] - Exercise 16 (A) [Page 196]

APPEARS IN

Selina Concise Mathematics [English] Class 9 ICSE
Chapter 15 Area Theorems [Proof and Use]
Exercise 16 (A) | Q 6 | Page 196

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