$ABCD$ is a line segment, trisected by the points $B$ and $C$. $P$ is any point on the circle where $BC$ is its diameter. If the angles $\angle APB$ and $\angle CPD$ are respectively $\alpha$ and $\beta$, prove that $$4\tan\alpha ⋅ \tanβ = 1$$
I tried dropping perpendiculars from $B$ and $C$ to $AP$ and $DP$ to get $tan\alpha$ and $tan\beta$ in terms of sides but I don't know how to utilise that. I also can't think of any other way to get $\tan\alpha$ and $\tan\beta$
On
Here is another approach:
Extend $PB$ to $E$ such that $PB=BE$.
Similarly extend $PC$ to $F$ such that $PC=CF$.
Join the lines as shown.
$(1)$ Note that $AB=BC$ and $PB=BE$ $\implies PAEC$ is a parallelogram.
$(2)$ Hence $AP // EC$ and $\alpha=x$.
$(3)$ Note that $\angle BPC=90^{\text o}.$
$(4)$ $$\therefore \tan \alpha = \tan x= \frac{PC}{PE} \tag{1}$$
$(5)$ Similarly $$\tan \beta = \tan y = \frac{PB}{PF} \tag{2}$$
$(6)$ From (4) and (5), $$\tan \alpha \times \tan \beta = \frac{PC \times PB}{PE \times PF}$$
$(7)$ $\because$ $PF= 2 \times PC$ and $PE = 2 \times PB$,
$(8)$ $$\therefore \tan \alpha \times \tan \beta = \frac{1}{4}$$
Since triangles $ABP$, $BCP$ and $CDP$ have common altitude from vertex $P$ and $AB = BC = CD$, these three triangles have same area: $$ \frac{1}{2}|AP|\cdot|BP|\sin\alpha = \frac{1}{2}|BP|\cdot|CP| = \frac{1}{2}|CP|\cdot|DP|\sin\beta $$ which implies $$ |BP| = |DP|\sin\beta, \quad |CP| = |AP|\sin\alpha. $$
On the other hand, we have $S_{APD} = S_{ABP} + S_{BCP} + S_{CDP} = 3S_{BCP}$: $$ \frac{1}{2}|AP|\cdot|DP|\sin\left(\frac{\pi}{2} + \alpha + \beta\right) = \frac{3}{2}|BP|\cdot|CP| = \frac{3}{2}|DP|\sin\beta\cdot|AP|\sin\alpha $$ which gives us $$ \sin\left(\frac{\pi}{2} + \alpha + \beta\right) = 3\sin\alpha\sin\beta $$ $$ \cos(\alpha + \beta) = 3\sin\alpha\sin\beta $$ $$ \cos\alpha\cos\beta = 4\sin\alpha\sin\beta $$ $$ 4\tan\alpha\tan\beta = 1 $$
Q.E.D.