A problem on tangent and secant lines

Question

My question is the following:

Assume a point $A$ outside some circle, and a point $X$ on the same circle, such that $AX$ is not a diameter of the circle. Draw $Y$ as the point of intersection between the circle and the (extesion of the) line $AX$. Draw the tangents to the circle in points $X$ and $Y$, and denote their point of intersection with $Z$. Also draw the two tangents $AB$, $AC$ to the circle, where $B$, $C$ are points on the circle. Prove that $Z$ lies on the extension of the line $BC$.

The approaches I have tried until now involved either

• analyzing triangle similarities and searching for equal angles (many pairs of angles subtend the same arcs),
• calculating cross ratios (considering the drawing as a two-point perspective drawing of a circle),
• establishing a particular origin of linear coordinates and searching vector expressions for the positions of all points,
• writing all points in polar coordinates,

although without any significant progress. Can you please help me formally solve this problem?

Thank you.

Answer 1

We're going to use two results on the polar(*) of a point with respect to a circle.

(1) if $P$ is exterior to a circle $\mathcal C, $ then the polar of $P$ is $(TT')$, where $T$ and $T'$ are the "tangent points";

(2) if $P$ and $M $ are two points different of the center of $\mathcal C$ , then $M$ belongs to the polar of $P$ iff $P$ belongs to the polar of $M$.

According to (1),

$\color{green}{BC}$ is the polar of $\color{blue}{A}$ and $\color{grey}{XY}$ the polar of $\color{red}{Z}$.

According to (2),

as $\color{blue}{A}$ is on $\color{grey}{XY}, \color{red}{Z}$ is on $\color{green}{BC}$. $\square $

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(*) To define the polar of a point $P$ in a circle $\mathcal C$ with center $O$ and radius $R$, you first determine $P' \in OP$ s.t.$$\overline{OP}.\overline{OP'}=R^2$$ then the polar is the perpendicular to $OP$ passing through $P'$.

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Let's Prove Quickly (1):

Let $P$ be a point outside the circle, $T$ the tangent point, and $H$ the foot of the perpendicular to $OP$ passing through $T$.

$OTP$ and $OTH$ are similar. So $\frac{OT}{OH}=\frac{OP}{OT}$. So $OP.OH=OT^2=R^2$. So $P'=H$ and $T$ belongs to the polar $p$ of $P$.$\square$

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Let's Prove Quickly (2): With the notations in the figure, it is assumed that$$P\in m$$with $m$, polar of $M$ Let $MH$ be perpendicular to $OP$. $OMH$ and $OM'P$ are similar. So $\frac{OH}{OM'}=\frac{OM}{OP}$. So $OP.OH=OM.OM'=R^2$. So $P'=H$ and $$M\in p$$where $p$ is the polar of $P$.$\square$

Answer 2

Building off of D S's comments (and the wikipedia article on harmonic quadrilaterals), we can see that $BXCY$ is harmonic.

However, if we let $C'$ be the other (one being $B$) intersection of ray $ZB$ and the circle, then $ZBC'$ is a line and $BXC'Y$ is harmonic. Since for fixed points $B,X,Y$ there is only one point $C$ on the circle such that $BXCY$ is harmonic (why?), meaning we must have $C=C'$ and thus $Z,B,C$ are collinear.

Answer 3

I feel like I should prove the properties (useful here) of the Harmonic quadrilateral stated in the Wikipedia article.

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We define a harmonic quadrilateral as a cyclic quadrilateral which has the product of opposite sides equal.

Lemma 1: Given three points $A,B,C$, and tangents $BT,CT$ to the circumcircle $(ABC)$, and $D = (ABC) \cap AT$, the quadrilateral $ABDC$ is harmonic.

Proof: By sine rule in $\Delta ABT$ and $\Delta ACT$, $$\frac{AB}{BT} = \frac{\sin(\angle ATB)}{\sin(\angle BAT)} = \frac{\sin(\angle DTB)}{\sin(\angle BAD)} \tag{1}$$ $$\frac{AC}{CT} = \frac{\sin(\angle ATC)}{\sin(\angle CAT)} = \frac{\sin(\angle DTC)}{\sin(\angle CAD)} \tag{2}$$ By sine rule in $\Delta DTB$ and $\Delta DCT$, $$\frac{DB}{TD} = \frac{\sin(\angle DTB)}{\sin(\angle TBD)} = \frac{\sin(\angle DTB)}{\sin(\angle BAD)} \tag{3}$$ $$\frac{DC}{TD} = \frac{\sin(\angle DTC)}{\sin(\angle TCD)} = \frac{\sin(\angle DTC)}{\sin(\angle CAD)} \tag{4}$$ Multiplying (1) and (4) and comparing with product of (2) and (3), we have $$\frac{AB\cdot DC}{BT\cdot TD} = \frac{\sin(\angle DTB) \sin(\angle DTC)}{\sin(\angle BAD)\sin(\angle CAD)} = \frac{AC \cdot DB}{CT \cdot TD}$$ Since $BT = CT$, we have $AB\cdot DC = AC \cdot DB$ as desired.

Lemma 2: In a harmonic quadrilateral $ABDC$, the tangents from $B$ and $C$ concur on $AD$ (this is just the converse of Lemma 1).

Proof: Let the two tangents be $BT$ and $CT$. Join $AT$ and let $AT \cap (ABDC) = D'$. This implies that $ABD'C$ is also harmonic, so that $$\frac{DB}{DC} = \frac{AB}{AC} = \frac{D'B}{D'C}$$ implying $$\frac{\sin(\angle DCB)}{\sin(\angle DBC)} = \frac{\sin(\angle D'CB)}{\sin(\angle D'BC)}$$This means that $D=D'$ since the function $f(X) = \frac{\sin(\angle XCB)}{\sin(\angle XBC)}$ on fixed arc $BC$ of a fixed circle is monotonic.
Instead of analyzing $f$, one could also provide a more elementary, albeit lengthy, proof using the fact that $\angle BDC = \angle BD'C$, and arguing that if $\sin(\angle DCB) < \sin(\angle D'CB) \iff \angle DCB < \angle D'CB$ also implies $\sin(\angle DBC)<\sin(\angle D'BC) \iff \angle DBC<\angle D'BC$. However, care must be taken to handle acute-obtuse configurations here.

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In your figure, lemma 1 implies $BXCY$ is harmonic, lemma 2 implies $Z$ lies on $BC$.