Pleas; Tangents Eillpse I tried to proof this issue, but no more satisfactory. Is it possible to get help or detailed solution Please

 A: Let $C_1,C_2$ be the foot of the perpendicular line from $F_1, F_2$ to the line $MA$ respectively. Also, Let $D_1, D_2$ be the foot of the perpendicular line from $F_1, F_2$ to the line $MB$ respectively. 
Then, what we want is $F_1C_1\times F_2C_2=F_1D_1\times F_2D_2$.
Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ (a\gt b\gt0)$. Also, let $e$ be the eccentricity of the ellipse. Then, we get $e=\frac{\sqrt{a^2-b^2}}{a}.$ And, we can write $F_1\ (ae,0),F_2\ (-ae,0)$.
Now letting $M(x_1,y_1)$, since each of the directrix is $x=a/e, x=-a/e$, we get $-a/e\lt x_1\lt a/e$, which is $a-ex_1\gt0, a+ex_1\gt0$.
Hence, we get 
$$F_1C_1=k(a-ex_1), F_2C_2=k(a+ex_1)$$
where $k=\frac{ab^2}{\sqrt{b^4{x_1}^2+a^4{y_1}^2}}.$
Hence, noting $b^2{x_1}^2+a^2{y_1}^2=a^2b^2$,
$$F_1C_1\times F_2C_2=k^2(a-ex_1)(a+ex_1)=\frac{a^2b^4(a^2-e^2{x_1}^2)}{b^4{x_1}^2+a^4{y_1}^2}=\frac{b^4(a^4-(a^2-b^2){x_1}^2)}{b^2(a^4-(a^2-b^2){x_1}^2)}=b^2.$$
We can get $F_1D_1\times F_2D_2=b^2$ as well. 
Thus, we can get $F_1C_1\times F_2C_2=F_1D_1\times F_2D_2$, which leads what you want. In the following, I'm going to prove it.
Let $F_1M=L_1, F_2M=L_2, \angle F_2MB=\alpha, \angle F_1MF_2=\beta, \angle F_1MA=\gamma.$
Then, since $F_1C_1\times F_2C_2=F_1D_1\times F_2D_2$, we have 
$$L_1\sin(\gamma)\times L_2\sin(\beta+\gamma)=L_1\sin(\alpha +\beta)\times L_2\sin(\alpha),$$ which is $-2\sin(\alpha+\beta)\times\sin(\alpha)=-2\sin(\gamma)\times\sin (\beta+\gamma)$.
By the well known formulas, 
$$\cos(2\alpha+\beta)-\cos(\beta)=\cos(\beta+2\gamma)-\cos(\beta).$$
Hence, $\cos(2\alpha+\beta)=\cos(\beta+2\gamma)$ leads $\cos(2\alpha+\beta)-\cos(\beta+2\gamma)=0$. 
Thus, we have
$$-2\sin(\alpha+\beta+\gamma)\sin(\alpha-\gamma)=0.$$
Since $0\lt \alpha+\beta+\gamma\lt\pi,-\pi\lt\alpha-\gamma\lt\pi$, we get $\alpha-\gamma=0$.
Now, we get what you want.
