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Given an angle $\angle ABC$ and a point $P$ inside of it, draw a circle which passed through $P$ and is tangent to both $\overleftrightarrow{AB}$ and $\overleftrightarrow{BC}$.

Well, the center has to lie on the angle bisector (obvious). The center must also lie on the perpendicular bisector of $P$ and a point $E$ on $\overleftrightarrow{AB}$ or $\overleftrightarrow{BD}$. Additionally, if $D$ is the center, then $\angle DEB = 90^\circ−\angle ABC$; in other words, $E$ is the intersection of the circle with diameter $BE$ and $AB$.

I've had little experience dealing with construction problems, so what are some "obvious" things to see?

https://www.geogebra.org/geometry/uzkujvb6

Thanks!

EDIT: I believe the problem is asking for a straightedge and compass construction.

Blue
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Baker013273213
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    Construct any circle tangent to the sides of the angle. Use $\overleftrightarrow{BP}$ to find a point $P'$ on that circle (there are two choices). Now, figure out how to construct the "matching" circle through $P$. – Blue Oct 17 '19 at 20:24
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    @Blue I think that’s probably the simplest straightedge-and-compass construction and would make a useful answer even without further details. – David K Oct 17 '19 at 20:35

2 Answers2

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Hint. Construct any circle tangent to the sides of the angle. Use $\overleftrightarrow{BP}$ to find a point $P'$ on that circle. (There are two choices.) Now, figure out how to construct the "matching" circle through $P$.

Note: Each choice of $P'$ yields a different matching circle.

Since OP has taken the hint, here's an illustration, with $\bigcirc O$ the "any circle":

enter image description here

Blue
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  • @Brethlosze: I stand by my answer. ... I guess by "Any point in the bisector is a possible circle", you mean "... possible center". That's true. So there are lots of candidates for the "any circle" I mentioned. Even so, the circles through $P$ resulting from the construction are fixed. – Blue Oct 17 '19 at 21:59
  • @Brethlosze: I don't want to give away everything, so I'll just say this: Determine a relationship between $P'$ and the center of its circle, and duplicate that relationship with $P$ to find the center of its circle. – Blue Oct 17 '19 at 22:25
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    @Blue OHHHH. Let the center of "any circle" be O, P' be the intersection of BP' and circle O, and the center of what we want is X. Since BP'O is similar to BPX, X is the intersection of BO and the line parallel to P'O from P. Right? – Baker013273213 Oct 18 '19 at 01:20
  • @Baker013273213: That's it! :) ... Since you've gotten the hint, I'll add an illustration to my answer shortly. – Blue Oct 18 '19 at 01:27
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Well, without more information about the construction, this is trivially done by drawing two parabolas with a ruler and a string, taking P as the foci, and each AB and BC as directrix.

The two intersections of these parabolas will be the two feasible centers.

Brethlosze
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