Hi mathlearn,

Considering your first question, it's given that $AY$ and $AX$ are congruent, making triangle $XYA$ isosceles. What then can you say about $\angle XYA$ and $\angle AXY$? Also, what's the relationship between angles $\angle AXY$ and $\angle BXE$?

As for your second question, consider triangle $CYE$. The measure of exterior angle $\angle YEB$ is equal to the sum of the measures of remote interior angles $\angle CYE$ and $\angle ECY$: $\measuredangle YEB = \measuredangle CYE + \measuredangle ECY$. It's given that $\angle ABC\cong \angle ACB$, and by a previous part $\measuredangle XYA = \measuredangle BXE$. By substitution you'll get $\measuredangle BEX = \measuredangle BXE + \measuredangle EBX$.

Hi mathlearn,

It does turn out in this case that $\overline{XE}$ and $\overline{AD}$ are perpendicular to $\overline{BC}$, but by corresponding angles, to show that $\overline{XE}\,\|\, \overline{AD}$, it is enough to show that $m\measuredangle BXE = m\measuredangle BAD$. Let $\alpha = m\measuredangle XYA$ and $\beta= m\measuredangle XBE$. By the remote interior angles theorem $m\measuredangle XAY = m\measuredangle XBE + m\measuredangle ACB = \beta + \beta = 2\beta$. Since the sum of the measures of the interior angles of triangle $XAY$ (the measures are $\alpha, \alpha$, and $2\beta$) is $180^\circ$, then $2\alpha + 2\beta = 180^\circ$. Thus, looking to triangle $ABC$, we find $m\measuredangle BAC = 2\beta$. Since $\overline{AD}$ bisects $\measuredangle BAC$, $m\measuredangle BAD = \alpha$. We also know that $m\measuredangle BXE = \alpha$ from a previous exercise, so $m\measuredangle BAD = m\measuredangle BXE$, as desired.

Now, if you want to prove perpendicularity of $\overline{XE}$ and $\overline{AD}$ (and hence that $\overline{XE}\, \| \,\overline{AD}$), use the fact that $\alpha + \beta = 90^\circ$ (which follows from the equation $2\alpha + 2\beta = 180^\circ$ above). Since $\alpha = m\measuredangle BXE$, looking to triangle $BXE$ we find $m\measuredangle XEB = 90^\circ$; as $\alpha = m\measuredangle BAD$ (shown above), looking to triangle $BAD$ we find $m\measuredangle ADB = 90^\circ$. Hence, $\overline{XE}$ and $\overline{AD}$ are perpendicular to $\overline{BC}$.