A geometry problem involving tangents of circumference

Question

I found out this problem on Geogebra but I can't prove it. Please help me. Thanks

Let $ABC$ be an acute triangle with circumference $(O)$. The tangents at $B, C$ of $(O)$ intersects at $D$. The tangent of $(O)$ at $A$ intersects the line $BC$ at $M$. $AD$ intersects $(O)$ at $E$. Now prove that $ME$ is the tangent of $(O)$ at $E$.

Answer 1

Hints towards a solution. If you're stuck, explain what you've tried.

The setup is quite complicated. How can we simplify it?

1. Point $E$ is quite troublesome, introduced only at the end, and not as relevant to the other points. Is there a way to remove it? Can we find a condition not involving $E$, that would imply that $ME$ is tangential?

If $OM$ is perpendicular to $AD$, then the tangents at $A$ and $E$ will meet at $M$.

2. How can we simplify this further? We see that either $A$ or $M$ is somewhat arbitrary (picking one forces the other). Since $A$ is sightly less involved in the setup (merely being a point on the circle), how can we "remove" $A$ from the above condition?

One approach to prove perpendicularity is to show that

$MA^2 - AO^2 = MD^2 - DO^2$

This approach allows us to "remove" $A$ via the substitutions $MA^2 = MB \times MC$ and $AO^2 = R^2$.

Now, we want to show that $MB \times MC - OB^2 = MD^2 - DO^2$. This becomes a condition only about $B, C$ and their common tangent $D$, and any point $M$ on line $BC$.

I assure you this is true, and again there are multiple ways of doing this (Pythagorean theorem, Coordinate geometry, brute force, etc), so pick an approach and push it through.

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Note: There are other ways of proving perpendicularity too. EG It's very easy (and arguably easier than recognizing the above) to coordinate geom / brute force the entire setup.

Answer 2

As said in the last comment of @Calvin Lin, an analytical geometry straightforward proof is possible.

Let us rephrase the issue into the following (equivalent) one : Let $(C)$ be a circle, $A,B$ two points on $(C)$, $D$ the intersection of the two tangents to the circle ; let $M$ be any point on line $AC$, taken outside the circle. Prove that the line joining the points of tangency of the tangents issued from $M$ passes through $D$.

(please note that we no longer need to give a special name to point $E$, because we don't need it, as in the answer by Calvin Lin).

Let us take a coordinate system where $B=(-1,0)$ and $C(1,0)$.

Let $O=(0,d)$ and $M=(m,0)$.

• It is rather easy to show that $D=(0,- \tfrac{1}{d})$ (you can for example use a classical relationship $h^2=aa'$ in right triangle $OBD$, where $h$ is the length of the altitude issued from $B$).

• the equation of circle $(C)$ is :

$$x^2+y^2-2dy-1=0\tag{1}$$

The equation of the polar line of point $(x_0,y_0)=(m,0)=M$ (which is the line joining the points of tangency of tangents issued from $M$) is deduced from (1) :

$$xx_0+yy_0-d(y+y_0)-1=0 \ \ \iff \ \ mx-dy-1=0\tag{2}$$

(See remark at the bottom).

It suffices now to check that the coordinates $(x,y)=(0,- \tfrac{1}{d})$ of $D$ verify equation (2).

Finished !

Remark about (2) :

• sanity check : if all zero indices are removed in (2), relationship (1) is retrieved.

• more deeply : the concept of polar line of a point with respect to a conic curve is central in projective geometry (deeply linked to the concept of duality). Very broadly speaking, it is like passing from a norm beginning by $x^2+y^2$ to a dot product beginning by $xx_0+yy_0$. Either you already know it, or you have to admit it for the present time, with the perspective to learn it later on...

This concept of polar line of a point with respect to a circle isa generalization of the concept of tangent at a point of the circle.