Calculate the intersection point of $m$ with $AC$.

Question

Let $ABC$ be the triangle with $A=(0,0)$, $B=(16,0)$ and $C=(4,16)$.

Let $B'$ be the midpoint of the side $\overline{AC}$, let $C'$ be the midpoint of the side $\overline{AB}$ and let $I$ be the point $(2\sqrt{17}-2, 9-\sqrt{17})$.

1. Calculate the line that goes through $B$ and $I$.

2. The angle bisector $m$ at $B$ of the triangle $ABC$ divides the opposite $\overline{AC}$ proportional to the adjacent sides. Calculate the intersection point of $m$ with $AC$. Show that this intersection point is on the line $BI$.

3. Show that $I$ has the same distance to $AB$ as to $AC$.

4. Conclude that $I$ is the center of the inscribed circle of the triangle $ABC$and calculate the line that goes through $I$ and the contact point of the inscribed circle and the side $\overline{BC}$.

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For question 1 we have:

\begin{equation*}y-0=\frac{9-\sqrt{17}-0}{2\sqrt{17}-2-16}(x-16) \Rightarrow y=-\frac{1}{2}(x-16) \end{equation*}

For question 2 we have:

We have to calculate the side lengths $|AB|$ and $|BC|$, right?

We have that $|AB|=16$ and $|BC|=\sqrt{12^2+16^2}=20$. Therefore $m$ divides the side $\overline{AC}$ in relation $\frac{20}{16}=\frac{5}{4}$, right? But how can we calculate the coordinates of that point?

For question 3 we have:

Here we just use the formula for distances and calculate both and see that these are equal, right?

Answer 1

What you need to do for part (2) is to compute the intersection point of the line you got from part (1), with the line that passes through $A$ and $C$, which you have not yet calculated. This is simply $$y = \frac{16}{4} x = 4x,$$ consequently the intersection point, which we will call $P$, satisfies the system $$\begin{align} y &= -\frac{x}{2} + 8 \\ y &= 4x. \end{align}$$ Then compute the distances $|PA|$ and $|PC|$, and show that because $$\frac{|PA|}{|AB|} = \frac{|PC|}{|CB|},$$ point $P$ obeys the proportionality criterion, thus $P$ lies on the angle bisector of $\angle B$. Conclude that, since this line is also the line passing through $BI$ from part (1), that $I$ lies on the angle bisector of $\angle B$.

For part (3), simply perform the distance calculation. The distance of $I$ to $AB$ is easy. However, the distance of $I$ to $AC$ requires us to find the line perpendicular to $AC$ passing through $I$, computing its intersection point $Q$ with $AC$, then computing the distance $|QI|$.

For part (4), observe from parts (2) and (3) that if $I$ lies on the angle bisector of $\angle B$, then the distance of $I$ to $BC$ equals the distance of $I$ to $AB$; therefore, $I$ is equidistant to all three sides of $\triangle ABC$, hence $I$ is the incenter.