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Suppose that there is line $l$ that is tangent to an ellipse $A$ at point $\,P\,$.

The ellipse has the foci $F'$ and $F$.

One then creates two lines - each from each focus to the tangency point $\,P\,$ .

What I want to prove is that the acute degree formed at $P$ between $l$ and the line segment $F'P$ equals the acute degree formed between $l$ and the line segment $FP$ .

How would I be able to prove this?

(ellipse has a horizontal axis as a major axis.) (from Proving a property of an ellipse and a tangent line of the ellipse)

Is there any way to do this without using trigonometry? I do understand the answers there, but I also need purely geometric ones without calculus, vector and trigonometry...

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1 Answer 1

Ok. I will prove that the external bisector of $\widehat{FPF'}$ is tangent to $A$ in $P$. Take $X$ on $FP$ such that $PX=PF'$ and $P$ lies between $F$ and $X$. In this way, $XF'$ is parallel to the internal angle bisector of $\widehat{FPF'}$. Call $r$ the external bisector of $\widehat{FPF'}$. Suppose that there is a point $Q\neq P$ on $r$ belonging to $A$:

$$ FX = FP+PF' = FQ + QF' = FQ + QX $$

must hold, but this is clearly impossible, since there is a unique point $P$ on $r$ such that $FP+PX=FX$. Since we got a contradiction, we have that the only point belonging to both $r$ and $A$ is $P$, so $r$ is tangent to $A$.

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