Cyclic Pentagon And Nice Angles

We have a cyclic pentagon, such that AE is a diameter and equals 10. We know BC = 5\sqrt{3} and \widehat{COD} = 18 .

Now F, G, H, I are the feet of the perpendiculars from C to AB, EB, AD, ED. If FG and HI concur at J, find FJI.

Now, let K be the foot of the perpendicular from C to AE. Notice that quadrilaterals CIDH (\widehat{CID} = \widehat{CHD} = \frac{\pi}{2} ) and CHKA (\widehat{CKA} = \widehat{CHA} = \frac{\pi}{2} ) are cyclic. Hence, we get:

\widehat{CHI} = \widehat{CDI} = \pi - \widehat{CDE} = \widehat{CAE} = \pi - \widehat{CHK}

This means that H, I, K are collinear. By an analogous reasoning, we get that even F, G, K are collinear.

Hence, FG, HI and AE concur at K, which actually is J.

Now, because AFCJ is cyclic (\widehat{CFA} = \widehat{CJA} = \frac{\pi}{2} ), we get \widehat{FJC} = \widehat{FAC} = \widehat{BAC} .

Similarly, from the ciclicity of CIEJ, we get \widehat{CJI} = \widehat{CED} .

Hence, we deduce \widehat{FJI} = \frac{\widehat{BOD}}{2} .

Because BD = 5\sqrt{3} and the radius of the circumference is 5, we conclude (by the sines theorem) that \widehat{BOD} = 138 .


\widehat{FJI} = 69

Scroll to top