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Byju's Answer

Standard IX

Mathematics

Converse of Theorem 2

A B C D is a ...

Question

ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC such that DF = 2 FC. Prove that AE CF is a parallelogram whose area is one third of the area of parallelogram AB CD.

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Solution

Given:

(1) ABCD is a parallelogram.

(2) E is a point on BA such that BE = 2EA

(3) F is a point on DC such that DF = 2FC.

To find:

Area of parallelogram

Proof: We have,

BE = 2EA and DF = 2FC

AB − AE = 2AE and DC − FC = 2FC

AB = 3AE and DC = 3FC

AE = AB and FC = DC

AE = FC [since AB = DC]

Thus, AE || FC such that AE = FC

Therefore AECF is a parallelogram.

Clearly, parallelograms ABCD and AECF have the same altitude and

AE = AB.

Therefore

Hence proved that

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ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC such that CE = 2 DE and F is the point of BC such that BF = 2 FC. Prove that:
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