The Triangle

Let $\angle{DAC} = \alpha$ and $\angle{ACP} = \beta$ , then $\angle{DPC} = \alpha+\beta$ . From the given condition, $\tan{ \alpha}=m$ , $\sin{ \beta}=1/a$ , $\tan{ (\alpha+\beta)}=m(1+a)$ . Hence, $\frac{1}{\sqrt{a^2-1}} = \tan{\beta} = \tan((\alpha+\beta) - \alpha) = \frac{m(1+a) - m}{1+m\cdot m(1+a)}=\frac{ma}{1+z}$ Therefore $z^2-(a^3-a^2-2)z+1 = (z+1)^2 - a^2(a-1)z =(z+1)^2 - a^2(a^2-1)m^2 = 0$

when we draw ABC as Isosceles triangle and make D a midpoint of BC and P a point on AD then PE perpendicular on AC it does make sense that ( AP/PD = BP/PE = a ) if only the triangle ABC is Equilateral ! in that case point P divides AD in way that AP:PD = 2:1 and also divides BE into BP:PE = 2:1 so (a = 2).

in ADB triangle we concluded that AB = 2DB and its Right-angled triangle in D

so => (AB)^2 = (AD)^2 + (DB)^2

4(DB)^2 = (AD)^2 + (DB)^2

3(BD)^2 = (AD)^2

(AD)^2/(DB)^2 = 3

since m^2 = (BD/AD)^2

then m^2 = 1/3

since z = 3m^2

then z = 3 x 1/3 = 1

then z^2 - (a^3 - a^2 - 2)z + 1 = 1 - (8 - 4 - 2)x1 + 1 = 0 !

have fun :)