5
$\begingroup$

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 $|PL|=|PK|$.

ThreeCircles

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?

Thank you for your help.

$\endgroup$
2
  • 2
    $\begingroup$ Please embed an illustrating diagram into your question, for clarity. For help embedding a diagram, see this article as well as this one. $\endgroup$ Sep 19 at 20:01
  • 1
    $\begingroup$ Yes. Provide a drawing and then you might get help. $\endgroup$
    – Moti
    Sep 20 at 6:13
1
$\begingroup$

Triangle ABC

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 $PK=PL$.

$\endgroup$
0
$\begingroup$

enter image description here

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:

enter image description here

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.

$\endgroup$
0
$\begingroup$

PointOnAngleBisector

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 $PK = PL$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.