Let $ABCDE$ be a cyclic pentagon inscribed in a circle of centre $O$ which has angles $\angle B=120^{\circ},\angle C=120^{\circ},$ $\angle D=130^{\circ},\angle E=100^{\circ}$. Show that the diagonals $BD$ and $CE$ meet at a point belonging to the diameter $AO$. Dinu Șerbănescu
Problem
Source: Romanian TST 2002
Tags: symmetry, trigonometry, geometry proposed, geometry
sunken rock
05.02.2011 19:34
Use the regular 18-sides polygon. One can further apply Ceva trigo form. Best regards, sunken rock
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JRD
05.01.2013 09:59
Hint : use Ceva theorem in sin mode in $\bigtriangleup ACD$
Aiscrim
27.01.2014 19:40
Let $AO\cap (O)=\{A^\prime \}$. It's well known that $AA^\prime\cap BD\cap CE \ne \emptyset \Leftrightarrow \dfrac{AB}{BC}\cdot \dfrac{CA^\prime}{A^\prime D}\cdot \dfrac{DE}{AE}=1$ But ${\dfrac{AB}{BC}\cdot \dfrac{CA^\prime}{A^\prime D}\cdot \dfrac{DE}{AE}=\dfrac{\sin{40^\circ}}{\sin{20^\circ}}\cdot \dfrac{\sin{30^\circ}}{\sin{10^\circ}}\cdot \dfrac{\sin{10^\circ}}{\sin{70^\circ}}=\dfrac{\sin{40^\circ}}{2\sin{20^\circ}\sin{70^\circ}}}=1$
soryn
10.02.2022 22:10
How to prove, please, the "well-known" above theorem?
Math4Life2020
10.02.2022 23:01
soryn wrote: How to prove, please, the "well-known" above theorem? Barycentric coordinates and/or vectors would work.
soryn
10.02.2022 23:39
I can't find any reference to the *well known"theorem...
Math4Life2020
11.02.2022 00:46
here you go: https://en.wikipedia.org/wiki/Ceva%27s_theorem
soryn
11.02.2022 08:29
Thank you, interesting...