# High school geometry proof help

I need help with a high school geometry proof. I think I've figured out why the prompt is true, but the proof attempt I've come up with seems very inelegant. Is there an easier method I'm missing?

Consider two circles with the second internally tangent to the first at point $A$ and also passing through the center of the first. Show that every chord of the first circle which has $A$ as an endpoint is bisected by the second circle.

My attempted proof:

Let there be any chord of the first circle which has $A$ as an endpoint. Let the other endpoint of the chord be called $B$.

Then let the following line segments be drawn:

• A segment connecting $A$ and the first circle's center $C$;
• A segment connecting $B$ and the first circle's center $C$; and
• A segment connecting the first circle's center $C$ with the point $I$ where the chord intersects the second circle.

Segments $AC$ and $BC$ are the same length because they both represent the radius of the first circle.

We have two right triangles $ACI$ and $BCI$. Since the two hypotenuses $AC$ and $BC$ are the same length and the two heights $CI$ are the same length, then the two bases $AI$ and $BI$ must also be the same length. Since $AI$ is half the length of the chord, the second circle bisects the chord.

• "We have two right triangles" does not root in something we know (without proof), assuming this immediately leads to establishing the claim. – dEmigOd Sep 28 '17 at 6:18
• Thanks for your reply. Would it be enough to reference the theorem stating that any triangle inscribed in a circle is a right triangle? – j.wood Sep 28 '17 at 6:23
• This is not true. Any triangle could be circumscribed – dEmigOd Sep 28 '17 at 6:25
• For me, the proof is good. But the theorem is: any triangle inscribed in a circle with one edge being its diameter is a right triangle – MMM Sep 28 '17 at 6:26
• So $AC$ should be established being a diameter of the smaller circle – dEmigOd Sep 28 '17 at 6:28

Another proof is to take the center of the circles being $C$ (the bigger), $D$ the smaller, the intersection point $I$ for the bigger circle and $J$ for the smaller.

We have that the triangles $ADJ$ and $ACI$ are similar since they have congruent angles (which is because they are isosceles and sharing an angle (of the leg). Also we have that $AD$ is half of $AC$ which implies that $AJ$ is half of $AI$.

We know that $D$ is on $AC$ because if it weren't we would have that $ADC$ would form a proper triangle and we would have that the smaller triangle doesn't tangent the larger. We also have that $D$ is the midpoint or otherwise the smaller circle wouldn't pass through $C$.

• I have posted a solution using 3D. – Jean Marie Dec 6 '17 at 8:51

Another proof: Since both CI and BD are perpendicular to AB, CI is parallel to BD. The fact that C is the midpoint of AD implies that I is the midpoint of AB.

• How do you know $CI \bot AC$? – dEmigOd Sep 28 '17 at 6:35
• @dEmigOd That's a known fact since $AC$ and $AD$ are diameters of respective circle. But you don't actually need to know that since you can use similar triangles instead. – skyking Sep 28 '17 at 6:39
• The fact is known if $AC$ is a diameter, how do you know it? – dEmigOd Sep 28 '17 at 6:40
• @dEmigOd Yes, but that's rather obvious. If it weren't we would be able to form a triangle $ACM$ with the midpoint of the smaller circle showing that it doesn't tangent the larger. – skyking Sep 28 '17 at 6:54
• @dEmigOd, You could consider a tangent to both circles at A. Because the angle between a tangent and a chord [AC] equals the angle subtended by that chord [AC] in the alternate segment (angle AIC), and the angle between a tangent and diameter [AD] is 90 degrees, therefore angle AIC is 90. – Kieren Pearson Oct 3 '17 at 1:31

From the tangency point $A$, blow up the small circle such that $$\text{The small circle} \rightarrow ^{\text{gets mapped to}} \text{The circle twice as big}$$

Since the dilation is of factor two, QED.

Though what I've said is a mere restatement of what others said before, the idea of dilation, called Homothety in general is an useful tool for Eucledian Geometry (check eg. the proof of the existence of Nine Point circle)

• [+1] Your solution is the most "economical" in thought and words. I have given another one that could as well been plainly described in terms of (3D) homothety ! – Jean Marie Oct 4 '17 at 14:56

I think next two facts give some hints.

1. In a circle chords, based on the same length arcs are of the same length
2. Arcs of the same angle are proportional to circle radius.

This immediately gives you $2AI=AB$, since those are chords on the arcs of the same angle in circles with radius ratio of $1:2$.

Here is an algebraic way. Let $r$ be the radius of the small circle and $R$ be the radius of the big circle. It's obvious that $R=2r$. Now, let $AB$ be the chord of the small circle and $AC$ be the chord of the big circle. Let point $A$ be the center of coordinate $(0,0)$, point $B(x_1,y_1)$, and point $C(x_2,y_2)$. The equation of small circle is \begin{align} x_1^2+(y_1-r)^2&=r^2\\ x_1^2+y_1^2&=2ry_1\tag1 \end{align} The equation of big circle is \begin{align} x_2^2+(y_2-2r)^2&=4r^2\\ x_2^2+y_2^2&=4ry_2\tag2 \end{align} From $(1)$ and $(2)$ we have $AB^2=2ry_1$ and $AC^2=4ry_2$. Then $$\frac{AB}{AC}=\sqrt{\frac{y_1}{2y_2}}$$ It's easily noticed that $\frac{y_1}{y_2}=\frac{1}{2}$, hence $$\frac{AB}{AC}=\frac{1}{2}$$

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One can use a trick that works rather often in 2D geometry, i.e., viewing it at as a 3D issue (see figure below).

Consider the big circle as being in a certain base plane. Let $A$ be the point of tangency of the two circles. Then consider the whole figure as the projection of a certain cone that is described in this way:

• apex $S$ is on the vertical axis issued from point $A$, at height $z=2$.

• it is cutted by plane $z=1$ along a "copy" of the small circle, that will be orthogonally projected onto the small circle.

• its generatrices are obtained by joining the apex $S$ to the generic point $M$ of the big base circle, with midpoint $M'$ situated on the copy of the small circle (for a clear reason of proportionnality: consider right triangle $SAM$).

As projections preserve the ratio of aligned points, we will still have $AM''=M''M$ ($M''$ denoting the vertical projection of $M'$).