Question

given that point g is the incenter of △hjk and that m∠dgh = 65°, what is m∠dhe? 65° none of these 50° 25°

Explanation:

Step1: Recall in - center property

The in - center is the point of intersection of the angle bisectors of a triangle. Since $G$ is the in - center of $\triangle HJK$, $GD\perp HJ$, $GE\perp HK$. In quadrilateral $DHEG$, $\angle GDH = 90^{\circ}$ and $\angle GEH=90^{\circ}$.

Step2: Use angle - sum property of a quadrilateral

The sum of the interior angles of a quadrilateral is $360^{\circ}$. Let $\angle DHE=x$ and $\angle DGH = 65^{\circ}$. Then, by the angle - sum property of a quadrilateral $ABCD$ ($\angle A+\angle B+\angle C+\angle D = 360^{\circ}$), in quadrilateral $DHEG$ we have $\angle GDH+\angle DHE+\angle GEH+\angle DGE=360^{\circ}$. Substituting $\angle GDH = 90^{\circ}$, $\angle GEH = 90^{\circ}$ and $\angle DGE = 65^{\circ}$ into the equation:
\[90^{\circ}+x + 90^{\circ}+65^{\circ}=360^{\circ}\]
\[x+245^{\circ}=360^{\circ}\]
\[x=\angle DHE=360^{\circ}- 245^{\circ}=115^{\circ}\] (This is wrong. Let's use another property).
Since $G$ is the in - center, $HG$ is the angle bisector of $\angle JHK$. In right - triangle $DGH$, $\angle DGH = 65^{\circ}$, $\angle GDH = 90^{\circ}$, so $\angle DHG=180^{\circ}-(90^{\circ}+65^{\circ}) = 25^{\circ}$. Since $HG$ is the angle bisector of $\angle JHK$, $\angle DHE = 2\angle DHG$. So $\angle DHE=50^{\circ}$.

Answer:

C. $50^{\circ}$