Question

In Figure, side AC of ABC produced to E so that CE = AC. D is the mid-point of BC and ED produced meets AB at F. RC, DP are drawn parallel to BA. Prove that FD = FE.
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Answer 1

Answer:

Step-by-step explanation:

ABC is a triangle.

D is midpoint of BC and DQ.

They're drawn parallel to BA.

Then,

Q is midpoint of AC.

AQ=DC

Since,

FA parallel to DQ∣∣PC.

AQC, is a transverse

So, AQ=QC and FDP also a transverse on them.

FD=DP (1) [ intercept theorem]

EC=1/2AC=QC

Now,

Triangle EQD,

Here C is midpoint of EQ and CP which is parallel to DQ.

And, P is midpoint of DE.

DP=PE(2)

Therefore,

From (1) and (2)

FD=DP=PE

FD=1/3FE

Hence, this is the answer.