$I$ is the incenter of $ABC$. The extension of $AI$ meets the circumcircle of $ABC$ at $D$. If $AB=3$, $AC=4$, and $[IBC]=[DBC]$, find $BC$.

Question

Question: Let $I$ be the incenter of $\triangle ABC$. The extension of $AI$ through $I$ meets the circumcircle of $\triangle ABC$ at $D$. If $AB=3$, $AC=4$, and $[\triangle IBC]=[\triangle DBC]$, then compute $BC$.

I initially thought that $IBDC$ was a parallelogram based on the diagram, but I later realised this is impossible because $IBDC$ would be a rhombus, implying that $\angle IEC=90^{\circ}$. However, this would also imply that $\angle ABD=\angle ACD=90^{\circ}$, which is clearly impossible because $AB \neq AC$.

Here's what I've done so far:

• Angle chasing yields $BD=ID=DC$.
• The condition $[\triangle IBC]=[\triangle DBC]$ implies that $IE=ED$ (the perpendiculars from $I$ and $D$ to $\overline{BC}$ have the same length and $\angle IEB=\angle DEC$)
• $ABDC$ is cyclic, so $\angle BAC=180^{\circ}-\angle BDC$. Using the law of cosines, it follows that $$\frac{25-BC^2}{3}+\frac{8x^2-BC^2}{x^2}=0$$ where $x=IE$.

Answer 1

Let $IE=x$ and $BC=7y$. You have rightly deduced that $$EI=DE=x,~~ BD=CD=2x.$$ Using the Angle Bisector Theorem in $\triangle ABC$, $$BE=3y,~~ CE=4y.$$ Using the same theorem in $\triangle ABE$, we get $AI=\frac{x}{y}.$

Using Ptolemey's Theorem, $$3\cdot 2x+4\cdot 2x=\left(\frac{x}{y}+2x\right)\cdot7y$$ Theorefore, $y=\frac{1}{2}\implies BC=\frac{7}{2}$.

Answer 2

I will denote by $a,b,c$ the sides of the triangle $\Delta ABC$, by $A,B,C$ (without a hat) its angles, by $S$ its area, by $h_a$ the height from $A$, by $d$ the common distance of the points $D,I$ to the side $BC$. So let us try to rewrite the given condition, which defines $d$, as an algebraic relation between the sides $a,b,c$: $$ \begin{aligned} \frac d{h_a} &= \frac{EI}{EA}=\frac{a}{a+b+c} \\ &\qquad \text{ shown for instance using Manelaus in $\Delta ABE$ w.r.t. $C$-bisector,} \\ &\qquad \text{ or by using the barycentrics $I=[a:b:c]$,} \\[3mm] \frac d{h_a} &= \frac{\frac a2\tan\frac {\widehat{CBD}}2}{h_a} = \frac{\frac a2\tan\frac A2}{\frac {2S}a} =\frac{a^2}{4S}\tan\frac A2 =\frac{a^2}{2bc\sin A}\tan\frac A2 \\ & =\frac{a^2}{4bc\sin \frac A2\cos\frac A2}\tan\frac A2 =\frac{a^2}{4bc\cos^2\frac A2} =\frac{a^2}{2bc(1+\cos A)} \\ &= \frac{a^2}{2bc\cdot\left(1+\frac{b^2+c^2-a^2}{2bc}\right)} = \frac{a^2}{(b+c)^2-a^2} = \frac {a^2}{(b+c+a)(b+c-a)}\ . \end{aligned} $$ The two expressions are equal, so $a=b+c-a$, i.e. $a=\frac 12(b+c)$.

In our case, the result thus involves a denominator, $c=3$, $b=4$, so $a=\frac 72$.

Picture check:

And indeed, changing the picture so that the constraint $b+c=2a$ is kept, does conserve the relation $IE=ED$.