Question

The side $AB$ of the parallelogram $ABCD$ is produced to $X$ and the bisector of $\angle CBX$ meets $DA$ produced and $DC$ produced at $E$ and $F$ respectively. Prove that $DE=DF=AB+BC$

Answer 1

Let $\angle CBX=x$

Given, $BF$ is the bisector of $\angle CBX$

$\therefore \angle CBF=\angle XBF=\frac{x}{2}$

$\angle ABE=\angle XBF=\frac{x}{2}$ ----Vertically opposite angles

$\angle EAB={180}^{\circ }-\angle DAB$ ----Sum of angles on a straight line is ${180}^{\circ }$

$={180}^{\circ }-x$

$\angle AEB={180}^{\circ }-[\angle EAB+\angle ABE]=\frac{x}{2}$

In $△AEB$, $\angle ABE=\angle AEB=\frac{x}{2}$ ----corresponding angles

$\therefore AE=AB$

Now we can prove that $BC=CF$

$DE=AD+AE$

$=AD+AB$ --------(i)

Similarly,

$DF=DC+CF$

$=DC+BC$ --------(ii)

$AD=BC$ and $AB=DC$ ---Opposite sides of a parallelogram

So, $DE=DF$

Hence, $DE=DF=AB+BC$ ---- from (i) and (ii)