Question

10) ABCD is a parallelogram and E is the midpoint of BC. The side DC is extended such that
it meets AE when extended at F. Prove that DF = 2DC.

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Answer 1

$\bold{GIVEN}$

ABCD is a parallelogram and E is the midpoint of BC, such that if AB and it is extended they meet at point F.

$\large {\boxed{\bigstar To \: Prove \: that \: DF = 2DC}}$

In ΔBAE and ΔEFC

$\boxed{ \mathsf{ \angle ABE = \angle ECF }}$

$\bold{REASON}$

Since, AD // BC (sides of parallelogram)

DAB and EBF are interior alternate angles.

$\boxed{ \mathsf{ \angle AEB = \angle CEF}}$

$\bold{REASON}$

Vertically opposite angles.

$\boxed{ \mathsf{ BE = EC}}$

$\bold{REASON}$

Since, E is the midpoint, it divides BC into equal parts.

$\therefore \boxed{ \triangle ABE \cong \triangle ECF }$ by AAS property.

So, by C. P. C. T. C $\boxed{AB = CF}$

Now,

$\bold{DF = DC + CF}$

Since,

CF = AB (As proved above)

AB = DC (Opposite sides of parallelogram)

$\therefore DC = CF$

So,

$\mathsf{DF = DC + DC}$

$\huge \boxed{\mathsf{DF = 2DC}}$

$\mathcal{HENCE \: PROVED}$

Answer 2

• Answer:

hope it helps........