Unexpected tangency

Question

I came across an "unexpected tangency" (for lack of a better term) while working out a different construction with GeoGebra.

The construction goes like this. Starting with a segment $AB$, draw a half-circle having $AB$ as its diameter. Let $P$ be the midpoint of this half-circle.

Pick a point $C$, different from $P$, but otherwise arbitrary, on the half circle. This defines a (rectangular) triangle $ABC$, with $AB$ as its hypotenuse.

Now, draw the (smallest) circle centered at $P$ that is tangent to one of the sides of the triangle $ABC$.¹

Next, let $D$ be the incenter of the triangle $ABC$, and draw the line perpendicular to $AB$ that goes through $D$.

To my surprise, this line is tangent to the circle centered at $P$.

Q. Can someone show me a proof of why this should be?

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BTW, the circle centered at $P$ is also tangent to the line passing through the other leg of the right triangle. Of course, this is apparent from the figure shown above, but one can prove it from the symmetry of the extended figure that results if one arranges the triangle $ABC$ together with three additional copies of it to form the pinwheel-like construction often seen in proofs of the Pythagorean theorem. I would love to see a simpler, less "operatic" proof of this fact.

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^{1In the figure shown, the side that the circle touches is $BC$, but in general, it will be either $BC$ or $AC$, depending on wich side of $P$ one picks for $C$.}

Answer 1

Let $R=AB/2$. Then (see figure below): $$ PE=PK\cos\theta=(R-OK)\cos\theta=(R-R\tan\theta)\cos\theta=R(\cos\theta-\sin\theta). $$ $$ OH=R-AH=R-(AC-DH)=R-2R\sin\theta+(R+OH)\tan{\theta\over2}, $$ whence: $$ OH=R{1-2\sin\theta+\tan{\theta\over2}\over1-\tan{\theta\over2}}. $$ Using the well known formulas $$ \sin\theta={2\tan{\theta\over2}\over1+\tan^2{\theta\over2}}, \quad \cos\theta={1-\tan^2{\theta\over2}\over1+\tan^2{\theta\over2}}, $$ one can check that $PE=OH$, as requested.

EDIT.

For the other tangency: $$ \begin{align} CE&=BC-BK-KE=2R\cos\theta-{R\over\cos\theta}-(R-OK)\sin\theta\\ &=2R\cos\theta-R\sin\theta-{R\over\cos\theta}+R\tan\theta\sin\theta= R(\cos\theta-\sin\theta)=PE \end{align} $$

Answer 2

Let $\triangle ABC$ have side-lengths $2a$, $2b$, $2c$, with $a>b$. Let the center of the (semi)circle, and midpont of $\overline{AB}$, be $O$; and let incircle $\bigcirc I$ touch $\overline{AB}$ at $T$.

We "know" (or can easily derive) that in any not-necessarily-right $\triangle ABC$, the length of the incircle-tangent segments from $B$ is given by $\frac12(|BC|-|CA|+|AB|)$; in this case, we have $|BT|= a-b+c$, so that $$|OT| = |BT|-|OB| = a-b \tag1$$

Now, let $M$ be the midpoint of $\overline{BC}$ (hence also the foot of the perpendicular from $O$), and let $\bigcirc P$, of radius $p$, touch $\overline BC$ at $Q$.

Since inscribed $\angle PCB = 45^\circ$, it follows that $\triangle PCQ$ is isosceles, so that $|QC|=p$. The perpendicular bisector of $\overline{CP}$ necessarily contains both $Q$ and $O$ (for different reasons), so that $\triangle OQM$ is also isosceles, and $|MQ| =b$. Therefore, $$a = b + p \quad\to\quad p = a-b = |OT| \tag2$$

Thus, $\overline{OT}$ is parallel and congruent to the "horizontal" radius of $\bigcirc P$, so that $\overleftrightarrow{IT}$ is tangent to $\bigcirc P$. $\square$

Answer 3

Intelligenti pauca's answer gave me an idea for a more visual proof. To make the proof independent of the specifics of the figure, I'll add the following to the problem statement: without loss of generality, assume that the vertices of the $ABC$ triangle are labeled such that $B$ denotes the vertex with the smallest angle. With this additional convention, we can assert that $|BC| > |CA|$.

Now for the proof, let the segment $EP$ be perpendicular to $AB$, and let $PT$ be the radial segment that is parallel to $AB$ (and, therefore, perpendicular to the line through the incenter $D$ and perpendicular to $AB$).

Following Intelligenti's pauca's strategy, I will show that $|PT| = |EF|$. From this it will follow that $T$ lies on the line through $D$ and perpendicular to $AB$. Since, by construction, the radial segment $PT$ is perpendicular to the latter, it will follow that $T$ is a point of tangency.

First, from the symmetry of the (extended) figure, it is clear that the diameter of the circle centered at $P$ is $|BC| - |CA|$. Therefore, $$|PT| = \frac{|BC| - |CA|}{2}.$$

Next, the segments connecting the incenter $D$ to the vertices $A$, $B$, and $C$ split the $ABC$ into three triangles. The heights of these triangles, all converging at $D$, further divide $ABC$ into six triangles. These six triangles can be grouped into three pairs, where the triangles in each pair share their hypotenuses. The triangles in such pairs are congruent.

Let $x$, $y$, and $z$ denote the lengths of the bases of these triangles, as shown in the figure.

Then we have that $|AB| = x + y$, $|BC| = y + z$, and $|CA| = z + x$. From this we get that

$$\frac{|BC| - |CA|}{2} = \frac{y - x}{2} = |PT|.$$

Lastly, point $E$ is the midpoint of the segment $AB$, since $P$ is the midpoint of the semicircle spanning $AB$. Therefore,

$$|EF| = \frac{|AB|}{2} - x = \frac{x + y}{2} - x = \frac{y - x}{2} = |PT|.$$