Let $\Gamma$ be a semicircle with diameter $AB$. On this diameter is selected a point $C$, and on the semicircle are selected points $D$ and $E$ so that $E$ lies between $B$ and $D$. It turned out that $\angle ACD = \angle ECB$. The intersection point of the tangents to $\Gamma$ at points $D$ and $E$ is denoted by $F$. Prove that $\angle EFD=\angle ACD+ \angle ECB$.
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Tags: geometry, semicircle, angles, equal angles
thedodecagon
21.09.2020 03:42
Let's say that $\angle D\Gamma E=x$. By the theorem to find the exterior angle between 2 secants (or tangents) (is there like a formal name for this?), we find that $\angle EFD=180-x$. Furthermore, we find that quadrilateral $\Gamma DFE$ is cyclic. if we can prove that $\angle DCE$ is on the circle containing $\Gamma DFE$, we can use the fact that inscribed angles having the same arc are equal (is this a formal name for this as well?) to prove that $\angle ECD=x$. And we should be done.
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collinator
21.09.2020 06:45
First we can extend this semicircle to be a full circle. The full circle is symmetric about line $AB$, since it is a diameter, and so if the reflections of $D$ and $E$ over $AB$ are $D'$ and $E'$ respectively, they will lie on this circle. Since $\angle DCA=\angle ECB$ the reflection of $E$ over $AB$ will lie on line $DC$, and the reflection of $D$ over $AB$ will lie on $EC$. Thus $DEE'D'$ is an isosceles trapezoid (since it is symmetric about diameter $AB$) with diagonals intersecting at $C$. $\angle DCE=\frac{1}{2}(\overarc{DE}+\overarc{D'E'})$ but since $DE=D'E'$ these arcs are congruent, so $\angle DCE=\overarc{DE}$ and since $\angle DFE=180-\overarc{DE}$ $\angle DFE=180-\angle DCE=\angle DCA+\angle ECB$. (Note that $\overarc{XY}$ here denotes minor arc $XY$)