In a parallelogram ABCD,AB = 10 , cm and AD = | vedclass

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(D) $ABCD$ is a parallelogram,in which $AB = 10 \, cm$ and $AD = 6 \, cm$. The bisector of $\angle A$ meets $DC$ in $E$. $AE$ and $BC$ produced meet a

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$4$

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(D) $ABCD$ is a parallelogram,in which $AB = 10 \, cm$ and $AD = 6 \, cm$. The bisector of $\angle A$ meets $DC$ in $E$. $AE$ and $BC$ produced meet at $F$.Since $AE$ is the bisector of $\angle A$,we have $\angle BAE = \angle EAD \quad \dots(1)$.Since $AD \parallel BC$ and $AF$ is a transversal,$\angle EAD = \angle EFB$ (Alternate interior angles) $\dots(2)$.From $(1)$ and $(2)$,$\angle BAE = \angle EFB$.In $	riangle ABF$,since $\angle BAE = \angle EFB$,the sides opposite to these angles are equal,so $BF = AB$.Given $AB = 10 \, cm$,therefore $BF = 10 \, cm$.We know $BF = BC + CF$. Since $BC = AD = 6 \, cm$ (opposite sides of a parallelogram are equal),$10 \, cm = 6 \, cm + CF$.$CF = 10 \, cm - 6 \, cm = 4 \, cm$.

In a parallelogram $ABCD$,$AB = 10 \, cm$ and $AD = 6 \, cm$. The bisector of $\angle A$ meets $DC$ in $E$. $AE$ and $BC$ produced meet at $F$. Find the length of $CF$ (in $cm$).

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