# Show: equality of angles in ellipse between foci and external point

first please take a loot at this:

Given is an ellipse with foci $F1, F2$ and an external point $P$. Through P I have constructed two tangents to the ellipse.

I need to show that: $\angle F1PB1 = \angle B2PF2$ and $\angle B1F2P = \angle PF2B2$. (ignore the angles around P, they are not meant to be seen here). This is what I tried so far:

• (For the second part I could show that $PF2$ is the bisectrix of $\angle UF2V$.)
• $U$ and $V$ can be created by reflecting $F1$ at $|PB1| and |PB2|$.
• Therefore I have $F1B1 = UB1$ and $F1B2 = VB2$ as well as $UP = F1P = VP$.

Any ideas, how to proof it?

• For the second part, see this page. – mathlove May 13 '16 at 16:23
• – Blue May 16 '16 at 8:33

It is well-known that the tangents $PB_1$ and $PB_2$ are the external angle bisectors of $\angle{F_1B_1F_2}$ and $\angle{F_2B_2F_1}$, so $F_2,B_1,U$ are collinear and $F_2,B_2,V$ are collinear.
From $PU=PF_1=PV$ and $F_2U=F_2B_1+B_1U=F_2B_1+B_1F_1=F_2B_1+B_2F_1=F_2B_1+B_2V=F_2V$ we can see that $PF_2$ is the perpendicular bisector of $UV$. Hence, $$\angle{F_2PU}=\angle{VPF_2} \\ \angle{F_2PB_1}+\angle{B_1PU} = \angle{VPB_2}+\angle{B_2PF_2} \\ \angle{F_2PB_1}+\angle{F_1PB_1} = \angle{B_2PF_1}+\angle{B_2PF_2} \\ \angle{F_2PB_1}+(\angle{B_2PB_1}-\angle{B_2PF_1}) = \angle{B_2PF_1}+(\angle{B_2PB_1}-\angle{F_2PB_1}) \\ \angle{F_2PB_1} = \angle{B_2PF_1}.$$