# Prove that $IL,JK$ and angle bisector of angle $BCD$ are concurrent

Given a convex quadrilateral $ABCD$. In $\Delta ABC$, $I$ is the incentre and $J$ is the excentre opposite to vertex $A$. Similarly, $K$ is the incentre and $L$ is the excentre opposite to vertex $A$ of $\Delta ACD$. Prove that the three lines $IL,JK$ and the angle bisector of angle $BCD$ are concurrent.

I tried assuming some angles in one if the triangle. Used some trigonometry, but the problem just got nasty. I don't think pure geometry would do it. The problem looks good. If possible, please solve using trigonometry and/or algebraic geometry. Thanks.

• Anyway, a good general technique for solving such concurrency problems is to use trilinear coordinates: to prove a concurrency boils down to proving that a determinant is zero, and computations are not that difficult. – Jack D'Aurizio Aug 21 '15 at 17:21
• The claim looks all right to me. The result holds for a cycluc quadrilateral. In a square, the concurrency point is $A$. – user167045 Aug 21 '15 at 17:43
• Also, the computation is difficult. Writing, the coordinates of the incentre and the excentre and then finding the line and the equation of the angle bisector and then substituting them in the determinant. Looks like a tough job to me. If it doesn't look to you, please post the solution. – user167045 Aug 21 '15 at 17:45

With somewhat-altered labels, we have this figure for $\square ABOC$ with incenters $P$ and $Q$ (and inradii $p$ and $q$) and excenters $R$ and $S$ (and exradii $r$ and $s$). We embed the figure in the coordinate plane with $O$ at the origin, and:

$$A = a\;(1,0) \qquad B = b\;(\cos 2\beta,\sin 2\beta) \qquad C = c\;(\cos 2\gamma,-\sin 2\gamma)$$ \begin{align} P &= \frac{p}{\sin\beta}(\phantom{-}\cos\beta,\sin\beta) \qquad Q = \frac{q}{\sin\gamma}(\phantom{-}\cos\gamma,-\sin\gamma) \\[4pt] R &= \frac{r}{\cos\beta}(-\sin\beta,\cos\beta) \qquad S = \frac{s}{\cos\gamma}(-\sin\gamma,-\cos\gamma) \end{align}

In/exradius formulas are known, and we can compute, for instance, $$\frac{p}{\sin\beta} = \frac{|\triangle OAB|}{\sin\beta\;(a+b+e)/2} = \frac{a b \sin 2\beta}{\sin\beta\;(a+b+e)} = \frac{2 a b\cos\beta}{a+b+e}$$ Likewise, $$\frac{q}{\sin\gamma} = \frac{2 a c\cos\gamma}{a+c+f} \qquad \frac{r}{\cos\beta} = \frac{2 a b \sin\beta}{a-b+e} \qquad \frac{s}{\cos\gamma} = \frac{2 a c \sin\gamma}{a-c+f}$$

From here, brute-force symbol manipulation (and, very likely, a slick geometric argument that has eluded me) determines the intersection point, $K$, of lines $\overleftrightarrow{PS}$ and $\overleftrightarrow{QR}$:

$$K = \frac{a b c}{a b + a c + c e + b f }\; \left(\;\cos 2 \beta + \cos 2 \gamma\;,\;\sin 2\beta - \sin 2\gamma\;\right) = \frac{a\;( c B + b C )}{a b + a c + c e + b f}$$

Since $c B + b C$ is a direction vector for the bisector of $\angle BOC$, we have that $K$ is on this bisector and is therefore the sought-after point of concurrency. $\square$

Edit. We can reduce some symbol manipulation, and get an interesting general concurrence criterion, by deferring the radius calculations. For instance, with \begin{align} P = p^\prime\;(\phantom{-}\cos\beta,\sin\beta) &\qquad Q = q^\prime \;(\phantom{-}\cos\gamma,-\sin\gamma) \\ R = r^\prime\;(-\sin\beta,\cos\beta) &\qquad S = s^\prime\;(-\sin\gamma,-\cos\gamma) \end{align} (where $p^\prime$, etc, are distances from $O$ along various angle bisectors, although not necessarily related to any in/exradii) we find that $$K = \frac{\cos(\beta+\gamma)}{p^\prime r^\prime + q^\prime s^\prime + (p^\prime q^\prime + r^\prime s^\prime ) \sin(\beta + \gamma)}\;\left(\;q^\prime r^\prime (P-S) + p^\prime s^\prime (Q - R) \;\right)$$ Ignoring the multiplied scalar, we force $K$ onto the bisector of $\angle BOC$ by substituting its components into the equation $y = x\,\tan(\beta-\gamma)$. The result is this relation:

$$q^\prime r^\prime s^\prime \cos\beta - p^\prime r^\prime s^\prime \cos\gamma - p^\prime q^\prime s^\prime \sin\beta + p^\prime q^\prime r^\prime \sin\gamma = 0$$

Or, better:

$$\frac{\cos\beta}{p^\prime} - \frac{\sin\beta}{r^\prime} = \frac{\cos\gamma}{q^\prime} - \frac{\sin\gamma}{s^\prime} \tag{\star}$$

(This seems like something that should have an elegant geometric derivation.) Substituting $p^\prime = 2 a b \cos\beta/(a+b+e)$, etc, causes each side of $(\star)$ to reduce to $1/a$.

• Nice. I just didn't get the last step. From the point $K$, how did you get to your RHS. Also, what is it you have written on the RHS?(it is new to me-$cB+bC$?) – user167045 Aug 23 '15 at 3:30
• @You-know-me: I'm (ab)using some vector shorthand. If $B = (x_1,y_1)$ and $C = (x_2,y_2)$, then $$c B + b C = c(x_1,y_1) + b(x_2,y_2) = (c x_1 + b x_2, c y_1 + b y_2)$$ Likewise, the "fraction" is just the vector $(cB+bC)$ multiplied by the scalar $a/(ab+ac+ce+bf)$. The last step is a matter of "noticing" that $K$ is a convenient combination of $B$ and $C$. That $cB+bC$ is the direction vector of the angle bisector follows from the fact that vector $B$ has length $b$, and $C$ has length $c$, so $cB$ and $bC$ have equal length $bc$; consequently $cB+bC$ is the diagonal of a rhombus. – Blue Aug 23 '15 at 3:43