Let $ABCD$ be a trapezoid, where $AD\parallel BC$, $BC<AD$, and $AB\cap DC=T$. A circle $k_1$ is inscribed in $\Delta BCT$ and a circle $k_2$ is an excircle for $\Delta ADT$ which is tangent to $AD$ (opposite to $T$). Prove that the tangent line to $k_1$ through $D$, different than $DC$, is parallel to the tangent line to $k_2$ through $B$, different than $BA$.
Problem
Source: VI International Festival of Young Mathematicians Sozopol, Theme for 10-12 grade
Tags: geometry, Parallel Lines
Assassino9931
23.09.2023 23:07
Bump (can anyone find a reasonably nice solution?)
VicKmath7
24.09.2023 00:40
Sure, here is a simple synthetic solution.
Let $I$ be the incenter of $\triangle TBC$, $J$ be the excenter of $\triangle TAD$; let the tangent from $D$ to $k_1$ meet $AB$ at $M$ and the tangent from $B$ to $k_2$ meet $CD$ at $N$. We will prove that $\angle TMD=\angle TBN$. We have that $\angle JID=90^{o}-\frac{\angle TMD}{2}$ and $\angle ABJ=90^{o}-\frac{\angle TBN}{2}$, so we need $\angle ABJ=\angle JID$. If $B'$ is the reflection of $B$ with respect to $TI$, we should prove that $IB'DJ$ is cyclic, which is true as $\angle TB'I=\angle TBI=\frac{\angle TBC} {2}=\frac{\angle TAD} {2}=\angle TJD$, as desired.
Comment: I just realized we can actually avoid constructing $B'$. The angle equality we should prove means that it's sufficient to prove that $\triangle TBJ \sim \triangle TID$, or that $\triangle TBI \sim \triangle TJD$, which is trivial by angle chasing.
Assassino9931
21.01.2025 13:24
Even though it's just similar triangles, the effective angle chasing (and simplifying the diagram) is far from immediate. Note that the angle between the second tangent to $k_1$ from $D$ and the line $CD$ is equal to $2\angle CDI$, while the angle between the second tangent to $k_2$ from $B$ and the line $AB$ is equal to $2\angle ABJ$ - hence the angle between this tangent and $CD$ equals $180^{\circ} - 2\angle ABJ - \angle ATD$. Hence it suffices to show that $2\angle IDC + (180^{\circ} - 2\angle ABJ - \angle ATD) = 180^{\circ}$, i.e. that $\angle ATD = 2\angle CDI - 2\angle ABJ$. Since $TJI$ is an angle bisector of $\angle ATD$, we reduce to $\angle ATJ + \angle ABJ = \angle CDI$, i.e. to $\angle BJT = \angle IDT$. Since $\angle BTJ = \angle DTI$ from the angle bisector, the latter is equivalent to $\triangle BJT \sim \triangle IDT$. Hence we could argue $\frac{BT}{TJ} = \frac{TI}{TD}$, i.e. $\frac{BT}{TI} = \frac{TJ}{TD}$, which would follow from $\triangle BIT \sim \triangle JDT$. In terms of angles, we have $\angle BTI = \angle DJT$ and $\angle TBI = \frac{1}{2}\angle ABC$, as well as $\angle DJT = 180^{\circ} - \left(90^{\circ} + \frac{1}{2}\angle ADT\right) - \frac{1}{2}\angle DTJ = \frac{1}{2}\angle DAT = \frac{1}{2}\angle ABC$, so we are done.