In the given figure, if D and E are the mid points of sides AB and AC, respectively, and DE is produced to F such that E is the mid point of FD, which of the following options is correct?

ΔAED ≅ ΔFEC

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ΔAED ≅ ΔCFE

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ΔADE ≅ ΔCEF

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ΔADE ≅ ΔCFE

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Solution

The correct option is D
ΔADE ≅ ΔCFE

In ΔADE and ΔCFE

DE = EF (E is the mid point of DF)

∠AED = ∠CEF (vertically opposite angles)

AE = CE (E is mid point of AC)

$∴$ ΔADE ≅ ΔCFE (by SAS congruency)

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In the given figure, if D and E are the mid points of sides AB and AC, respectively, and DE is produced to F such that E is the mid point of FD, which of the following options is correct?

The correct option is D ΔADE ≅ ΔCFE In ΔADE and ΔCFE DE = EF (E is the mid point of DF) ∠AED = ∠CEF (vertically opposite angles) AE = CE (E is mid point of AC) ∴ ΔADE ≅ ΔCFE (by SAS congruency)

The correct option is D
ΔADE ≅ ΔCFE

In ΔADE and ΔCFE

DE = EF (E is the mid point of DF)

∠AED = ∠CEF (vertically opposite angles)

AE = CE (E is mid point of AC)

$∴$ ΔADE ≅ ΔCFE (by SAS congruency)