HELP A synthetic geometry problem involving the inscribed circle of a triangle

Question

PROBLEM:

Let $ABC$ be a triangle with $BC>AB$. We denote by $D,E,F$ the contact points of the inscribed circle with $BC,CA,AB$ and with $I$ the center of this circle. Let $K=AD \bigcup BI$ and $P= KF \bigcup BC$. If the point $Q$ is symmetric to $A$ with respect to $E$, prove that $PQ \perp CI$.

WHAT I THOUGHT OF

DRAWING:

Okey, so the first thing that came in my mind when i read the text is to write that because $BC>AB$, $\angle BAC > \angle ACB$.

$AD,BE,CF$ are the bisectors of the angles of the triangle. We know that the center of a inscribed circle is the intersections of the bisectors.

Then, i don't understand the intersections. In my drawing $BI$ intersects $AD$ in $I$ which means that $I$ is the same as $K$. The same thing for $P= KF \bigcup BC$.

Is there my drawing wrong or the way i see the intersectons wrong? Hope one of you can help me. Thank you.

Answer 1

$\underline{\bf{\color{red}{Note}}}$:$\space$ We provide below an answer/proof. But we have a request. Before going through this answer, you need to finish the conversation you have started with @Lynn to clear your doubts about the geometrical configuration.

To facilitate our proof, we added the line segments $DF$to the diagram that portraits the scenario describe in OP’s problem statement. For brevity, we let $\measuredangle ABC=2\beta$, $\measuredangle BCA=2\omega$, and $\measuredangle FKA=\alpha$.

From the properties of the incircle of a triangle, we know that $IB$ and $IC$ are the angle bisectors of $\angle ABC$ and $\angle BCA$ respectively and these lead to, $$\measuredangle ABI=\measuredangle IBC=\beta \quad\text{and}\tag{1a}$$ $$\measuredangle BCI=\measuredangle ICA=\omega. \tag{1b}$$

We can also deduce that, $$ID=IE=IF=r, \tag{1c}$$ $$AE=AF,\tag{1d}$$ $$BD-BF, \quad\text{and}\tag{1e}$$ $$CD=CE.\tag{1f}$$

Because of (1a) and (1e), according to SAS congruence rule, the triangle pair $BDK$ and $BKF$ are congruent. So, we have, $$KD=KF \quad\text{and}\tag{2a}$$ $$\measuredangle BDK=\measuredangle KFB. \tag{2b}$$

Since both angle pairs ($\angle BDK,\space\angle KDP$) and ($\angle KFB,\space\angle AFK$), it follows from (2b) that, $$\measuredangle KDP=\measuredangle AFK. \tag{3}$$

Furthermore, since $\angle FKA$ and $\angle PKD$ are opposite angles, we shall write, $$\measuredangle FKA=\measuredangle PKD=\alpha. \tag{4}$$

Using (2b), (3), and (4), and quoting ASA congruence rule, we can state that the triangle pair $FKA$ and $PKD$ are congruent. Hence, we have, $$AF=DP. \tag{5}$$

It is given that $AE=EC$. Using (1d) and (5), we can infer that, $$EC=DP.\tag{6}$$

We can use (1f) and (6) to show that, $CP=CQ$, which makes the $\triangle PCQ$ isosceles. Since $CI$ is the angle bisector of the apex angle of the isosceles tringle $PCQ$, it is perpendicular to the base $PQ$ of the said triangle.

Answer 2

You found the incenter by drawing and intersecting the bisectors, which is good, but you incorrectly took e.g. the point where the bisector through $A$ intersects $BC$ and called that $D$.

However, $D$ is supposed to be the point where the inscribed circle contacts $BC$, and in practice these are not going to be the same.

In this diagram I took from https://mathworld.wolfram.com/AngleBisector.html you can see $T_1$ is not the point where the inscribed circle touches the right side $A_2A_3$:

I tried adding an inscribed circle to your drawing and showing how the real $D,E,F$ will differ from the points you called $D,E,F$.

I encourage you to make a more precise drawing — $BE$ looks pretty off in particular — to convince yourself that this error does not merely result from artistic imprecision.

Here is a sketch in GeoGebra that includes $K$, $P$, and $Q$: