$ABCD$ is a parallelogram, $AD$ is produced to $E$ so that $DE = DC = AD$ and $EC$ produced meets $AB$ produced in $F$. Prove that $BF = BC$.

Given:

$ABCD$ is a parallelogram, $AD$ is produced to $E$ so that $DE = DC = AD$ and $EC$ produced meets $AB$ produced in $F$.

To do:

We have to prove that $BF = BC$.

Solution:

From the figure,

In $\triangle ACE$,

$O$ and $D$ are the mid points of sides $AC$ and $AE$.

This implies,

$DO \parallel EC$ and $DB \parallel FC$

$BD \parallel EF$

Therefore,

$AB = BF$

$AB = DC$                           (Opposite sides of a parallelogram are equal)

This implies,

$DC = BF$

In $\triangle EDC$ and $\triangle CBF$,

$DC = BF$

$\angle EDC = \angle CBF$                ($\angle EDC = \angle DAB$ and $\angle DAB=\angle CBF$ corresponding angles)

$\angle ECD = \angle CFB$        (Corresponding angles)

Therefore, by ASA axiom,

$\triangle EDC \cong \triangle CBF$

This implies,

$DE = BC$             (CPCT)

$DC = BC$

$AB = BC$

$BF = BC$               (Since $AB = BF$)

Hence proved.

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Updated on: 10-Oct-2022

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