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In the given figure, ABCD is a parallelogram, E is a point on BC and the diagonal BD intersects AE at F.

Prove that DF × FE = FB × FA.

In the given figure, ABCD is a parallelogram, E is a point on BC and the diagonal BD intersects AE at F. Similarity of Triangles, RSA Mathematics Solutions ICSE Class 10.

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Answer

Since, ABCD is a || gm

∴ AD || BC

In ΔADF and ΔEBF,

∠ADF = ∠EBF [Alternate angles are equal]

∠AFD = ∠EFB [Vertically opposite angles are equal]

∴ ΔADF ∼ ΔEBF (By A.A. axiom)

Corresponding sides of similar triangles are proportional.

DFFB=FAFE\dfrac{DF}{FB} = \dfrac{FA}{FE}

DF × FE = FB × FA

Hence, proved that DF × FE = FB × FA.

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