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Question 14
D and E are the mid-points of the sides AB and AC, respectively, of ΔABC.DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is
(a) DAE=EFC
(b) AE=EF​​​​​​​
(c) DE=EF​​​​​​​
(d) ADE=ECF​​​​​​​
 


Solution

In ΔADE and ΔCFC, suppose DE = EF

Now,   AE = CE    [Since, E is the mid-point of AC]

Suppose  DE = EF 

And AED=FEC   [Vertically opposite angles]

ΔADEΔCFE   [ By SAS congruence rule]

AD = CF    [By CPTC rule]

And ADE=CFE  [ By CPCT]

Hence,    AD ∥ CF   [ Since, alternate interior angles are equal]

Therefore, we need an additional information which id DE = EF

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