# Prove that if the line is parallel, then the line segment has a maximum length

Two circles with centres $O_1$ and $O_2$ intersect in points $A$ and $B$. The line $MP$ goes through points $M$, $A$ and $P$ ($M$ and $P$ are two other intersection points).
How can I prove that MP will have a maximum length if $MP$ is parallel to the line $O_1O_2$?

• This question is missing some information. It looks like the line has to go through one of the circle intersections. Is that true? – amd Apr 17 '17 at 17:43
• The line MP goes through the point A. (it is the line MAP) @amd – idliketodothis Apr 17 '17 at 17:49

The quadrilateral made by $O_1,O_2$, the midpoint of $AM$ and the midpoint of $AP$ is a right trapezoid, hence the length of $MP$ is at most twice the length of $O_1 O_2$, unless the previous trapezoid is indeed a rectangle, i.e. iff $MP\parallel O_1 O_2$.

• @idliketodothis: because in a circle the perpendicular bisector of a chord always goes through the center of the circle. I am updating with a picture. – Jack D'Aurizio Apr 17 '17 at 18:12
• And how did you decide that the length of MP is the most twice as the length of O1O2? Have never heard of such a theorem. – idliketodothis Apr 17 '17 at 18:18
• @idliketodothis: let $JK$ be the segments having as endpoints the midpoints of $AM$ and $AP$. Then $JK\leq O_1 O_2$ because we have a right trapezoid and $MP=2\,JK$ because that is obvious, hence $MP\leq 2\,O_1 O_2$. – Jack D'Aurizio Apr 17 '17 at 18:30