Question

In a ΔABC, the line joining the mid points D and E of sides AB and AC respectively, is produced to meet at point F such that DE = EF. What can you say about the figure DFCB?

A

Rhombus

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B

Parallelogram

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C

Kite

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D

Rectangle

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Solution

The correct option is B Parallelogram Consider the triangles ΔADE and ΔCFE DE = EF (given) ∠AED = ∠CEF (vertically opposite angles) AE = CE (E is the mid point of AC) Thus, ΔADE ≅ CFE (by SAS congruency) ∴ AD = CF (By CPCT) But, AD = DB (D is the mid point of AB) ∴ DB = CF (1) Also, ∠DAE = ∠FCE (By CPCT) ∴ AB || CF (Alternate angles are equal) (2) Hence, from (1) and (2), we have, DB || CF and DB = CF Thus, DFCB is a parallelogram.

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