euclidean-geometrygeometrytriangles
Problem:
In the below figure, $E$ is the excenter of $\triangle ABC$ such that $\angle ACB=5x.$ Also, point $F$ is on $AC$ such that $\angle CBF=x$ and point $D$ is on the extension of $BC$ such that $DE\perp AE,\; DE\perp BH,\;$ where $H$ is the intersection of $DE$ and $BF$. Find the largest integer value of $x.$
My progress:
I marked all the angles I found but they all result in the same equation
$2\alpha =180^{\circ} – 5x \implies \boxed{\alpha = 90^{\circ} – \frac{5x}{2}}$
When I see that $\angle CAB$ is partitioned as $24^\circ$ and $18^\circ$, what angle is left to make an $60^\circ$ angle (which is for making an equilateral triangle, which gives us more equal sides)? After doing that, we see that $\angle ECB = 36^\circ$ and this makes $\angle ABE = 30^\circ$. Now, notice that $\Delta ADB$ is congruent to $\Delta AEB$. So we have $|AD| = |AE| = |AC|$. So the answer is $78^\circ$.
Sometimes, geometry consists of dropping the right line or introducing the accurate point...
Denote by $D$ the point on $AP$ such that $\angle PDC=90°$ and let $BP=x$. Note now that the triangle $\Delta PCD$ is a $30°-60°-90°$ triangle, which implies that $PD=x$.
Therefore $\Delta BPD$ is isosceles and hence $\angle PBD=30° \Rightarrow \angle DBA=15°$.
Easy angle chasing leads to the conclusion that $DC=DB=DA$, which implies that $\Delta CDA$ is also isosceles. Thus $\angle ACD=45°$. $$\therefore \angle ACB=45°+30°=75° \blacksquare$$
Best Answer
Observe, $$\angle BHD=\angle AED=90^{\circ}\implies AE\parallel BH$$ $$\;\angle EAF=\angle BFA=\angle FBC+\angle FCB=6x$$ Extend ray $BA$ to some point $P$. Since $E$ is the ex-center and $AE\parallel BH$, $$\angle FBA=\angle EAP=\angle EAC=6x$$ Therefore, in $\triangle ABC$, $\angle ABC=7x\;$ and $\;\angle ACB=5x$. $$\angle ABC+\angle ACB<180^{\circ}\implies 12x<180^{\circ}\implies x<15^{\circ} $$ The largest possible integer value of $x$ is $14^{\circ}$. In order to show that that $x=14^{\circ}$ is possible, chase angles and note that all given constraints are satisfied.