Let $ABC$ be a triangle with $AB<AC$. Let $D$ be on the side $AC$ such that $AD=AB$. Let $I$ be the incenter of triangle $ABC$ and $E$ be the intersection of the perpendicular bisectors of $ID$ and $BC$. Then find $DE$ in terms of $AB$, $AC$ and $BC$.
In the original question, some lengths had been given ($AB=5$, $AC=8$ and $BC=7$). So it made my calculations much easier. I could not get any elegant solution so I had to use coordinate geometry to get the solution with the brute-force method. An interesting thing I observed was that $AI$, the perpendicular bisector of $BC$ and perpendicular bisector of $ID$ met at the circumcircle of $\triangle ABC$. I know why $AI$ and the perpendicular bisector of $BC$ meet at the circumcircle but I do not know why the perpendicular bisector of $ID$ would also meet there. Could anyone please provide proof?
Back to the main question. I set $B=(0,0)$ and $C=(7,0)$ and without much difficulty, I found the coordinates of $A=\left(\frac{5}{7}, \frac{20 \sqrt{3} }{7} \right)$ and $I=(2, \sqrt{3})$ and $D=\left( \frac{65}{14} ,\frac{15\sqrt{3}}{14}\right)$. The equation of the perpendicular bisector of $BC$ is $x=3.5$ and equation of perpendicular bisector of $ID$ is $y-\frac{29\sqrt{3}}{28}=\frac{-37}{\sqrt{3}} \left(x-\frac{93}{28}\right)$. Their intersection is the point $E= \left(\frac{7}{2}, \frac{-7\sqrt{3}}{6}\right)$. And hence the length of $DE$ is $\frac{7\sqrt{3}}{3}$.
As you can already see this method is extremely cumbersome and prone to mistakes (I do not even know if my answer is correct). I would really love a geometric solution to this question. I only know elementary geometry and trigonometry, so I request a solution based on those. (Also please clarify why they will be collinear). Any solution will be appreciated. Please feel free to use the lengths given to ease your calculation and writing.
Diagram for reference:
2 Answers
$\begingroup$I'll denote the angles of the triangle by $A,B,C$ and sides of the triangle opposite to these angles respectively as $a,b,c$.
Fact 1: Since $\triangle ABD$ is isosceles with angle-bisector $AI$, $AI$ is the perpendicular bisector of $BD$.
Fact 2: $$\begin{aligned}\angle BIC&=180^\circ-\angle IBC-\angle ICB=180^\circ-\dfrac{B}2-\dfrac{C}2\\&=180^\circ-\left(\dfrac{B+C}2\right)=180^\circ-\left(\dfrac{180^\circ-A}2\right)=90^\circ+\dfrac{A}2 \end{aligned}$$
Fact 3: Using that $\triangle ABD$ is isosceles, we have $\angle ABD=\angle ADB= \dfrac{180^\circ-A}2$ $$\begin{aligned} \angle BDC= A+\angle ABD=A+\left(\dfrac{180^\circ-A}2\right)=90^\circ+\dfrac{A}2\end{aligned}$$
From Facts 2,3, we have that $\angle BIC=\angle BDC$, so if we draw the circumcircle of $\triangle BDC$, then these two equal angles are angles in the same segment of that circle, so that $B,I,D,C$ are concyclic. This implies that $\underline{\text{$\triangle BID$ and $\triangle BDC$ have the same circumcircle}}$.
Consequently, denoting $\texttt{PB,CC}$ as perpendicular bisector and circumcircle respectively, it follows that $$(\texttt{PB} \text{ of } ID) \cap (\texttt{PB} \text{ of } BC)= \texttt{CC} \text{ of } \triangle BIC \ (\text{by definition of }\texttt{CC})\\ (\texttt{PB} \text{ of } BD) \cap (\texttt{PB} \text{ of } BC)= \texttt{CC} \text{ of } \triangle BDC \ (\text{by definition of }\texttt{CC}) $$ but the underlined fact implies $$\texttt{CC} \text{ of } \triangle BDC=\texttt{CC} \text{ of } \triangle BIC$$ so that $$(\texttt{PB} \text{ of } ID) \cap (\texttt{PB} \text{ of } BC)=(\texttt{PB} \text{ of } BD) \cap (\texttt{PB} \text{ of } BC)$$and by Fact 1 $$(\texttt{PB} \text{ of } ID) \cap (\texttt{PB} \text{ of } BC)=AI \cap (\texttt{PB} \text{ of } BC)=E$$ where the second equality is something you mentioned you already know.
Thus $E$ is the center of circumcircle of $\triangle BIC,\triangle BDC$ with $EC,EI,ED$ as radius $R_1$ of the circle. You can find $|DE|=R_1$ using the sine rule on $\triangle BDC$ as
$$\begin{aligned} DE=IE=R_1=\dfrac{BC}{2\sin(\angle BDC)}&=\dfrac{BC}{2\sin\left(90^\circ+\dfrac{A}2\right)}\\&=\dfrac{a}{2\cos\left(\dfrac{A}2\right)}\end{aligned}$$and then use $\cos(A/2)=\sqrt{\dfrac{1+\cos^2A}2}$ with $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$ from the cosine rule.
$\endgroup$ 3 $\begingroup$We can see that $ID=IB$ . Also we have:
$\angle BEA=\angle ACB$
$\angle BAE=\angle EAC$
Which results in:
$\triangle ABE=\triangle AEC $
Therefore:
$BE=DE$
Now connect C to I. CI is bisector of angle ACB. Mark intersection of perpendicular from E on ID as F. This perpendicular is also the bisector of angle IED. We have:
$\angle IEF=\frac{\angle BEI}{2}=\frac{\angle ACB}{2}$
Which results:
$\angle BCI=\angle IEF$
We can also show that $EC=ED$. This means points B, I, D and C are on circumference of a circle center at E. Now consider triangle IBC , DE is in fact the radius of this circle, so we use following formula to calculate it's measure:
$DE=R=\frac{BI\times IC\times BC }{2S}$
Where S is the area of this triangle. Now all we need is to find the measures of BI, IC. Having these we can find S using Harum's formula and find R.
$\endgroup$ 6
