anglegeometrytriangles
In an isoceles triangle ABC the angle ABC equals 135 degrees. The point F lies on AC. The angle ABF equals 90 degrees. AG is the bisector of the angle BAC, where G is the intersection point of BC and the bisector. How can I find the value of the angle GFC?
It's quite easy to find the angle A, but what shall I do next? I tried finding many angles, finding sums of the triangle's angles to find the value of GFC, but it was worthless.
let angles *=y and o=x, AE=DE so base angles are equal. and by CPCT angle EDC will be 2y. So from triangle ADC y+y+2y+x=180 and from triangle ABC x+x+y+y+x=180, solving these x=36 and hence 2x=72 degrees
The point where circle $\alpha$ intersects $AB$ is the foot of the perpendicular from $P$ on $AB$, call this point $D$. Circle $\beta$ is tangent to $\alpha$ at point $D$ and thereafter the midpoint of $AP$, $D$, and the centre of $\beta$ are collinear. By similarity, the centre of $\beta$ lies on the line parallel to $AP$ through the point $B$. Extend $PD$ to meet this parallel line at point $Q$. $BQ$ is the diameter of circle $\beta$. Now, observe that $K$ is the foot of the perpendicular from $Q$ on $BC$.
So we reframe the problem as follows :
In $\triangle ABC$, the internal angle bisector of $\angle A$ is drawn and two parallel lines $l_{1}$ and $l_{2}$ to this line are drawn through $B$ and $C$ respectively. $P$ is any point on the angle bisector. Perpendiculars from $P$ on $AB$ and $AC$ intersect $AB$ and $AC$ at $D$ and $E$ respectively. When extended, $PD$ and $PE$ meet $l_{1}$ and $l_{2}$ at $Q$ and $R$ respectively. Perpendiculars $QK$ and $RL$ on $BC$ are drawn intersecting $BC$ at $K$ and $L$ respectively. Prove that, $PK=PL$.
Draw parallel lines to $AB$ and $AC$ through $P$ intersecting $BC$ at $M$ and $N$ respectively.
Now, $\frac {PM}{PN}=\frac {AB}{AC}$.
$PQ=AB\cdot\frac {PD}{AD}$ ($\triangle PDA\sim \triangle QDB$) and similarly $PR=AC\cdot\frac {PE}{AE}=AC\cdot\frac {PD}{AD}$.
$\Rightarrow \frac {PQ}{PR}=\frac {AB}{AC}=\frac {PM}{PN}$
$\angle MPQ=\angle NPR=90^{\circ}$ and hence $\triangle QPM\sim \triangle RPN$.
Quadrilaterals $KQPM$ and $NRPL$ are cyclic. So, $\angle PKM=\angle PQM=\angle PRL=\angle PNL$ and therefore $PK=PL$.
Best Answer
Well, it's simple. If you calculate the various angles around vertex $B$ you realize that $BC$ is the exterior angle bisector of triangle $ABF$ through vertex $B$. Also, $AG$ is the angle bisector of $\angle\, BAC = \angle \, BAF$. Since the exterior angle bisectors of vertices $B$ and $F$ of triangle $ABF$ and the interior angle bisector of its vertex $A$ must intersect at a common point, that common point is $G = AG \cap BC$. Hence, $FG$ is the exterior angle bisector of $ABF$ through $F$. Thus, $$\angle \, GFC = \frac{1}{2} \,\angle BFC = \frac{1}{2} \,(\angle \, BAF + \angle \, ABF) = \frac{1}{2} \left(\frac{45^{\circ}}{2} + 90^{\circ}\right)$$