Let $A,B,C$ and $D$ be $4$ points in the plane such that any combination of three or more of them are non-collinear. Let $[AB]$ and $[CD]$ intersect at $M$. Suppose that: $AM = 6$, $MB = 4$ and $MD = 3$.
What should be the length of $MC$ so that $A,B, C$ and $D$ are on the same circle?
It is the first time that I encounter such a problem. I constructed a figure and located the center by finding the intersection of the perpendicular bisectors of the chords $[AB]$ and $[CD]$, but I think that this won't get me anywhere.
Any ideas?
Thanks guys.
A quadrilateral is cyclic (i.e. there exists a circle that can be circumscribed around it) if and only if (iff) its opposite angles sum up to $180^{\circ}$.
This is straightforward to see and I believe you already know this, but it can be seen by observing that by the inscribed angle theorem an inscribed angle is half the corresponding central angle that subtends the same arc.
All arcs sum up to $360^{\circ}$ and so the opposite inscribed angles will sum up to $180^{\circ}$. The 'if' part can be proved easily too (assume a circle goes through $A,B,C$ but does not go through $D$ and instead intersects $CD$ at $D'$. This will give you a contradiction (due to the 'only if' part) in both cases when $D'$ is on the segment $CD$ and when $D'$ is on the ray $CD$ but not on the segment $CD$).
$$ABCD\text{ is cyclic}\iff\angle A=\angle BDM\iff \triangle MAC\sim\triangle MDB\iff \frac{x}{AM}=\frac{MB}{MD}$$
where $x:=MC$.