On a cyclic quadrilateral $ABCD$, there is a point $P$ on side $AD$ such that the triangle $CDP$ and the quadrilateral $ABCP$ have equal perimeters and equal areas. Prove that two sides of $ABCD$ have equal lengths.
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Tags: geometry, cyclic quadrilateral, perimeter, PMO, 2010
16.11.2015 13:27
17.11.2015 19:03
Posted Before geometry
02.11.2016 05:14
03.12.2020 22:41
By equal perimeters, $$PD+CD=PA+AB+BC\quad (1)$$Calculating area of $ABCD$ and $PDC$, knowing that $S_{PDC}=\frac{1}{2}\cdot S_{ABCD}$, we get that $$S_{ABCD}=AB\cdot BC\cdot \frac{1}{2}\cdot\sin{\angle ABC}+DA\cdot DC\cdot \frac{1}{2}\cdot\sin{\angle PDC}=\frac{1}{2}\cdot\sin{\angle ABC} (AB\cdot BC+DA\cdot DC)$$$$CD\cdot DP\cdot \frac{1}{2}\cdot\sin{\angle PDC}=S_{PDC}=\frac{1}{2}\cdot S_{ABCD}$$By the last two we obtain that $$2\cdot CD\cdot DP =AB\cdot BC+DA\cdot DC\implies AB\cdot BC+CD\cdot AP=DP\cdot DC$$Multiplying $(1)$ by $CD$, we get $$AB\cdot BC+CD\cdot AP=PA\cdot DC+AB\cdot DC+BC\cdot DC-CD\cdot DC\implies AB\cdot BC+CD^2=AB\cdot DC+BC\cdot DC$$The last is equivalent to $$BC\cdot (AB-CD)= DC\cdot (AB-CD),$$thus $AB=CD$ or $BC=CD$. We are done!