Angle Quadrisector

Let the four equal parts be $\angle ACD$ , $\angle DCE$ , $\angle ECF$ , $\angle FCB$ , where ADEFB is a straight line. It is obvious that $CE$ is the angle bisector. WLOG let $AC>CB$ . Then $CF$ is the perpendicular and $CD$ is the median. Let $\angle ACD=\alpha$ , $EF=FB=b$ , and $AD=DB=c$ . Then $\tan \alpha =b$ , $\tan 2\alpha=c-b$ , $\tan 3\alpha=2c-b$ . Thus, $\tan 3 \alpha = 2c-b = 2(c-b)+b = 2 tan 2\alpha + \tan \alpha$ . Using the double and triple angle formula for tangent, we obtain $\frac {3b-c^3}{1-3b^2} = \frac {4b}{1-b^2}+b$ . Clearing denominators, we obtain $b(b^4 - 6b^2 + 1 )=0$ . If $b = 0$ , then $\tan \alpha = 0 \Rightarrow \alpha = 0$ , which makes it a degenerate triangle. Thus, $b^4-6b^2 + 1 = 0 \Rightarrow b^4 - 2 b^2 + 1 = 4 b^2 \Rightarrow \frac {2b}{1-b^2} = \pm 1$ . Notice that this is $\tan 2 \alpha = \pm 1$ , hence we either have $2 \alpha = 45^\circ$ or $135 ^\circ$ (we reject the latter). Hence, $4 \alpha = 90^\circ$ .

Let $D, E$ and $F$ be the intersection of $AB$ with the perpendicular, the angle bisector and the median of $C$ , respectively. Since the angle bisector splits the angle into 2 equal parts, we may assume that the perpendicular is closer to $A$ (else flip the vertices $A, B$ ), so points $A, D, E, F$ and $B$ lie in order.

Since $\angle ADC = \angle CDE = 90^\circ$ and $\angle ACD = \angle ECD$ , we have that triangles $CAD$ and $CED$ are congruent. Thus $CE = AC$ and $ED=DA$ . By the angle bisector theorem, we have $\frac {CA}{CB} = \frac {AE}{EB}$ , $\frac {CB}{CE} = \frac {BF}{FE}$ and $\frac {CF}{CD}=\frac {FE}{ED}$ .

Let $AD = x$ and $BF = y$ , so $y = BF = AF > AE = 2x$ and $FE = y-2x$ . Then $\frac {2y-2x}{2x} = \frac {BE}{EA} = \frac {BC}{CA} = \frac {BC}{CE} = \frac {BF}{FE} = \frac {y}{y-2x}$ . This gives $y^2 -4xy + 2x^2 =0$ . Taking the equation as a quadratic in $y$ and using the quadratic formula, we have $y = (2 + \sqrt{2} )x$ (ignore other root since $y > 2x$ .)

Since $\angle CDF = 90^\circ$ , we have $\angle CFD = \sin^{-1} \left(\frac {CD}{CF}\right)$ and $\frac{CD}{CF} = \frac {DE}{EF} = \frac {x}{y-2x} = \frac {1}{\sqrt{2} }$ . So $\angle CFD = \sin^{-1} \left( \frac{1}{\sqrt{2}}\right) = 45^\circ$ which implies that $\angle FCD = 180^\circ - 90^\circ - 45^\circ = 45^\circ$ and thus we have $\angle ACB = 2\times 45^\circ = 90^\circ$ .

Let $D,E$ and $F$ be the points on $AB$ such that $CD$ , $CE$ and $CF$ are respectively the altitude, angle bisector and median of the vertex $C$ . WLOG, let $AC < BC$ . Since they split $\angle ACB$ into 4 equal parts, let $\angle ACD = \angle DCE = \angle ECF = \angle FCB = \alpha$ .

In the two right-angled triangles $ACD$ and $CDB$ , $\angle ACD = \alpha$ and $\angle BCD = 3\alpha$ and, therefore, their complementary angles in the respective triangles are $\angle A = 90^\circ - \alpha$ and $\angle B = 90^\circ - 3\alpha$ .

Applying the law of sines to $\triangle ACB$ , we have

\displaystyle \begin{aligned} \frac{AC}{CB} &= \frac{\sin B}{\sin A}\\ & \\ &= \frac{\sin (90^\circ - \alpha)}{\sin (90^\circ - 3\alpha)}\\ & \\ &= \frac{\cos \alpha}{\cos 3\alpha}. \tag{0} \end{aligned}

On the other hand, we can apply the law of sines to $\triangle ACF$ and $\triangle CFB$ to get another result for $\displaystyle \frac{AC}{CB}$ .

\displaystyle \frac{AC}{\sin \angle CFA} = \frac{AF}{\sin 3\alpha}\tag{1}

and

\displaystyle \frac{CB}{\sin\angle CFB} = \frac{AF}{\sin 3\alpha}\tag{2}

respectively. Here, we note $\angle CFA = 180^\circ - \angle CFB$ and $\sin \angle CFA = \sin (180^\circ - \angle CFB) = \sin \angle CFB$ . So, dividing (1) through (2) yields

\displaystyle \frac{AC}{CB} = \frac{\sin 3 \alpha}{\sin \alpha}. \tag{3}

Equating (0) and (3) yields

$\displaystyle \frac{\sin 3 \alpha}{\sin \alpha} = \frac{\cos \alpha}{\cos 3\alpha}$

which implies $\displaystyle \sin 3\alpha \cos 3 \alpha = \sin \alpha \cos \alpha$ . Using $\sin 2 \theta = 2 \cos \theta \sin \theta$ , it is equivalent to $\displaystyle \sin 6\alpha = \sin 2 \alpha$ . Since $0<4\alpha < 180^\circ$ , we see that $6\alpha < 360^\circ$ . So we have either $6 \alpha = 2 \alpha$ or $6 \alpha = 180^\circ - 2\alpha$ . The former must be rejected for $\alpha$ cannot be zero, and the latter gives $\alpha = 22.5^\circ$ which is consistent with $4\alpha < 180^\circ$ . So, $\angle ACB = 4\alpha = 90^\circ$ .

Here, let CD, CE and CF be the perpendicular, angle bisector and median respectively. Now the angle is split into 4 equal parts. So angle bisector must be between perpendicular and median. Hence, CE is between CD and CF. Now let the measure of angle ACB be 4Y. So angle of each division = Y. Now, we know that a median divides a triangle into two triangles of equal area. Hence, ar.(CFB) = ar.(CAF) where ar.() denotes area of figure ar.(CFB) = ar.(CAD) + ar.(CDE) + ar.(CEF) Applying sine rule regarding area of triangle ½ BC *CF sinY = ½ CF CE sinY + ½ CE CD sinY +1/2 CD CA sinY implies BC CF = CF CE+CE CD+CD CA ...........(1) Now, CD is the perpendicular. So, CDE, CDF, CAD and CDB are right triangles and applying trigonometry, CE = CD secY, CF = CD sec(2Y), CB = CD sec(3Y), CA= CD secY Putting these values in equation ....(1) we get {CD sec(3Y)} * {CD sec(2Y)} = {CD sec(2Y)} {CD sec(Y)} + {CD} {CD secY} + {CD} {CD sec(Y)} implies sec(3Y)* sec(2Y) = sec(Y)* sec(2Y) + 2 sec(Y) Solving this equation for Y we get {sin(Y)}^2 = (2+ (2)^0.5)/4 or {sin(Y)}^2 = (2-(2)^0.5)/4 \Rightarrow Y = 67.5 or Y = 22.5 where angles are in degrees But if Y = 67.5 then 4Y = 270 , So this case is not possible in a triangle Therefore Y = 22.5 And hence angle ACB = 4Y = 4 22.5 = 90

that's 22.5 x 4 = 90