# For a triangle $ABC$, prove that for a point $P$ on the angle bisector of $\angle BAC$,$|PL|=|PK|$ where $L,K$ are intersections of circles with sides

In a triangle $$ABC$$, the point $$P$$ lies on the angle bisector of $$\angle BAC$$. We define a circle $$\alpha$$ with diameter $$AP$$, circle $$\beta$$ which passes through $$B$$ and is tangent to $$\alpha$$ at the point $$E$$, where it intersects $$AB$$, and circle $$\gamma$$ which passes through $$C$$ and is tangent to $$\alpha$$ at the point $$F$$, where it intersects $$AC$$. Then the points $$K,L$$ are defined as the intersections of $$BC$$ with $$\beta$$ and $$\gamma$$ respectively. The task is to prove that $$|PL|=|PK|$$.

In the case when $$ABC$$ is isosceles with $$CB$$ as the base, we can see that since $$\alpha$$ passes through $$A$$ and $$P$$ which lies on the angle bisector, it passes through $$AC$$ and $$AB$$ symmetrically, i.e. the distances from $$A$$ to $$D$$ (intersection of $$\alpha$$ and $$AB$$) and from $$A$$ to $$E$$ (intersection of $$\alpha$$ and $$AC$$) are the same. Then from the fact that the trinagle is isosceles, $$\beta$$ and $$\gamma$$ intersect $$CB$$ symmetrically, i.e. $$|CL| = |KB|$$. Hence, for $$KLP$$, the angle bisector of $$\angle BAC$$ is also the angle bisector of $$\angle KPL$$. $$KLP$$ is then similar to $$ABC$$ and therefore isosceles with $$LK$$ as base.

I have not been successful with generalising this proof to all triangles, however. It certainly cannot make use of symmetry in a similar manner. How should I approach this?

• Please embed an illustrating diagram into your question, for clarity. For help embedding a diagram, see this article as well as this one. Sep 19 at 20:01
• Yes. Provide a drawing and then you might get help.
– Moti
Sep 20 at 6:13

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 $$PK=PL$$.

Hint: M is mid point of KL. You have to show trapezoids TPMK and MPNL are equal which results in PL=PK.

You may also use following figure:

In this figure M and N are circomcircles of triangles APK and APL respectively. AE is radical axis of these circles and EL=EB. If DE=KB then DK=EL and triangles PKD and OLE are equal that means PL=PK.

We reproduce the diagram given with the problem statement after extending it by adding the line segments $$GB$$, $$GK$$, $$GD$$, $$HC$$, $$HL$$, and $$HD$$, where $$D$$, $$G$$, and $$H$$ are the centers of the circles $$\alpha$$, $$\beta$$, and $$\gamma$$ respectively. We also draw the segments $$KE$$ and $$LF$$, and then extend both so that they meet at $$N$$. Please note that the line passing through $$N$$ and $$P$$, which intersects with $$BC$$ at $$M$$, is also added.

For brevity, let $$\measuredangle CAB = 2\phi$$. Furthermore, we have $$\measuredangle BAP = \measuredangle PAC = \phi$$, because $$AP$$ is the angle bisector of the angle $$CAB$$.

Let’s do some angle chasing. Since $$DE$$, $$DA$$, and $$DF$$ are radii of the circle $$\alpha$$, the triangles $$EDA$$ and $$ADF$$ are isosceles triangles. As a consequence, $$\measuredangle AED = \measuredangle DFA = \phi. \tag{1}$$

Since, the angles $$BEG$$ and $$HFC$$ are vertically opposite to the angles $$AED$$ and $$DFA$$ respectively, we have from (1), $$\measuredangle BEG = \measuredangle HFC = \phi. \tag{2}$$

Since $$GB$$ and $$GE$$ are radii of the circle $$\beta$$, the triangles $$EGB$$ is an isosceles triangle. Therefore, using (2), we shall write, $$\measuredangle GBE = \measuredangle BEG = \phi. \tag{3}$$

Since $$HC$$ and $$HF$$ are radii of the circle $$\gamma$$, the triangles $$CHF$$ is an isosceles triangle. Therefore, as per (2), we get, $$\measuredangle FCH = \measuredangle HFC = \phi. \tag{4}$$

Now, pay attention to the circle $$\beta$$ exclusively. According to (3), the angle on the other side of the apex angle of the isosceles triangle $$EGB$$ can be expressed as $$\measuredangle BGE = 180^o + \measuredangle GBE + \measuredangle BEG =180^o + 2\phi. \tag{5}$$

Angles $$BKE$$ and $$BGE$$ are subtended by the same arc of the circle $$\beta$$ at its circumference and the center respectively. From (5), it follows that $$\measuredangle BKE = \frac{\measuredangle BGE}{2} = 90^o + \phi. \tag{6}$$

The angle $$EKL$$ is the supplement of the angle $$BKE$$. Hence, from (6), we have, $$\measuredangle EKL = 90^o - \phi. \tag{7}$$

It is time to divert our attention to the circle $$\gamma$$. Using (4), the angle $$CHF$$ at the apex of the isosceles triangle $$CHF$$ can be written as $$\measuredangle CHF = 180^o - \measuredangle FCH - \measuredangle HFC =180^o - 2\phi. \tag{8}$$

Angles $$CLF$$ and $$CHF$$ are subtended by the same arc of the circle $$\gamma$$ at its circumference and the center respectively. From (8), it follows that $$\measuredangle CLF = \frac{\measuredangle CHF}{2} = 90^o - \phi. \tag{9}$$

According to (7) and (9), $$\triangle LNK$$ is an isosceles triangle. Its apex angle $$LNK$$ is given by $$\measuredangle LNK = 180^o - \measuredangle NKL - \measuredangle KLN = 2\phi. \tag{10}$$

Therefore, we have $$\measuredangle FNE = \measuredangle FAE$$, which means that the point $$N$$ lies on the circle $$\alpha$$. Since $$PNE$$ and $$PAE$$ are angles in the same segment, $$\measuredangle PNE = \measuredangle PAE = \phi$$. Due to a similar reason, $$\measuredangle FNP = \measuredangle FAP = \phi$$ as well. Now, we have enough evidence, i.e. $$\measuredangle PNE = \measuredangle FNP$$, to state that $$NM$$ is the bisector of the angle $$LNK$$. Since $$\triangle LNK$$ is an isosceles triangle, $$NM$$ is perpendicular to $$KL$$, which makes the point $$M$$ the midpoint of the segment $$KL$$.

As the final step of this proof, consider the two right triangles $$KMP$$ and $$PML$$, where we have $$\measuredangle KMP = \measuredangle PML = 90^o$$. The side $$MP$$ is common to both triangles. Since $$M$$ is the midpoint of $$KL$$, we have $$KM =ML$$. According to SAS rule, the two triangles are congruent requiring $$PK = PL$$.