Let $ABCD$ be a convex cyclic quadrilateral. Prove that: $a)$ the number of points on the circumcircle of $ABCD$, like $M$, such that $\frac{MA}{MB}=\frac{MD}{MC}$ is $4$. $b)$ The diagonals of the quadrilateral which is made with these points are perpendicular to each other.
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Tags: geometry, circumcircle, angle bisector, geometry proposed
numbertheorist17
07.04.2013 06:14
Let $R$ be the circumradius of $ABCD$.
Note that since \[\dfrac{MA}{MB}=\dfrac{MD}{MC},\] $MA\cdot MC=MD\cdot MB$. Note that \[\dfrac{[MAC]}{[MBD]}=\dfrac{\dfrac{MA\cdot MC\cdot AC}{4R}}{\dfrac{MB\cdot MD\cdot BD}{4R}}=\dfrac{AC}{BD}.\] Thus, the altitudes from $M$ to $AC$ and $BD$ have the same length. Let $T$ be the intersection of $AC$ and $BD$. Then, $M$ lies on the angle bisector of one of the angles $ATB$, $BTC$, $CTD$, or $DTA$. Checking shows that all 4 points satisfy the condition.
Also, note that the diagonals intersect at $T$ by the above observation, and thus are perpendicular.
Mahdi_Mashayekhi
09.01.2022 13:37
MA.MC = MB.MD ---> [AMC]/[BMD] = AC/BD so M has equal distances from AC and BD so M lies on angle bisector of angle created by AC and BD and there are 4 angles so we have 4 points like M note that this angle bisectors are in fact diagonals of this quadrilateral and are perpendicular to each other.
AgentC
31.12.2023 22:20
what does [Triangle's name] mean?