Bisector Intersection Proof

Question

$AGF$ is a triangle, $B$ is the center of its inscribed circle, and $D$ and $E$ are the intersections of that circle with $[AG]$ & $[AF]$ respectively. I couldn’t figure out how to prove that the perpendicular bisector of $[AD]$, the perpendicular bisector of $[AE]$, and the angular bisector of $\widehat{GAF}$ intersect. ($H$ on the figure). Do you know how to prove this intersection? Thank you in advance!

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https://www.geogebra.org/m/fhge9wjr

Answer 1

Triangles $\color{brown} {ADB}$ and $\color{red}{AEB}$ are equal (two equal angles and a common side).

Therefore $$AE=AD$$ Consider the symmetry $s$ with respect to $AB$. $$s(D)=E$$

Let $d$, the perpendicular bisector of $[AD]$ and $e$ , the perpendicular bisector of $[AE]$ $$s(d)=e$$ So $d, e$ and and the angular bisector of $\widehat{GAF}$ intersect.

Answer 2

One proof is that perpendicular bisectors of $AD$ and $AE$ meet at the circumcenter of $\triangle{ADE}$. Therefore $HD=HE$, (and $AD=AE$)*, and all points equidistant from two lines that meet at a point fall on the angle bisector of those lines.

• for DS: since $DB=BE$ and $AB$ is shared and $\angle{ADB}, \angle{AEB}$ are right angles, by HL theorem $\triangle{ADB}=\triangle{AEB}$, $AD=DE$, i.e., $ADBE$ is a kite.