tag:blogger.com,1999:blog-6933544261975483399.post6471416561928158157..comments2024-04-22T04:55:16.794-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 39Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-8196651865163430012012-03-31T22:33:31.614-07:002012-03-31T22:33:31.614-07:00http://img805.imageshack.us/img805/5227/problem39....http://img805.imageshack.us/img805/5227/problem39.png<br />see attached sketch and shorter solution for question 1 and 2 below<br />∠ (HOC)= ∠A/2+∠C/2 = 90-∠B/2<br />Triangle FBD is isosceles => ∠ (BDF)= ∠(HDC)= 90-∠B/2<br />So ∠ (HOC)= ∠ (HDC) => OCHC Cyclic<br />Similarly AGFO is cyclicPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59616283988790414052009-05-14T11:37:00.000-07:002009-05-14T11:37:00.000-07:00It seems that the statements are closely related. ...It seems that the statements are closely related. I will try:<br /><br />(1)We need some definitions:<br />The point of intersection of CG and AB will be called K.<br />The point of intersection of FD and BO will be called L.<br />Observe that BFOD is a kite (two tangent rays, two radii), and thus FLO=90°.<br />Now chase some angles: AOC=180°-A/2-C/2=90°+B/2 (triangle ACO). Then GOA is supplementary, ergo 90°-B/2. Thus in triangle AOG, the third angle AKO=90°-A/2+B/2. Its supplementary angle is OKB=90°+A/2-B/2. Then in triangle KOB the last angle is BOK=90°-A/2. Now we have the triangle OLG with LOG=BOK=90°-A/2 and GLO=FLO=90°. Then the third angle is OGL=A/2.<br /><br />So we have OGF=OGL=A/2=OAF, and therefore AGFO is cyclic.<br /><br />(2) is similar.<br /><br />(3) As AFO=OEA=90°, A,F,O and E lie a common circle, which is the same one as around the quadrangle AGFO. So AGFE is cyclic.<br /><br />(4) is similar with (3)<br /><br />(5) The three angles are inscribed angles in the circle around A,G,F,O,E.<br /><br />(6) is similar with (5).<br /><br />(7) AGO=AFO=90°, as, according to (1), the four points lie on a common circle. (The other relation is similar.)<br /><br />(8) ECDO is a kite (two tangent rays, two radii), whence the diagonals CO and ED are orthogonal. But according to (7) the line through CO is also orthogonal to AG. So ED and AG are parallel. (The other relation is similar).<br /><br />(9)CAB=EAF=EGF=EGH. (Again, the other relation is similar.)<br /><br />(10) Follows directly from (9), as there are already two pairs of equal angles in ACB and GEH.<br /><br />(11) Follows directly from (9), again with the two pairs of equal angles.<br /><br />(12) Follows from (7) and and the converse of Thales' Theorem.<br /><br />(13) All angles are inscribed angles in the circle through A,G,F,O and E on the chord EO.<br /><br />(14) is similar with (13).<br /><br />(15) We have shown OGH(=OGL)=A/2 during the proof of (1) and EGH=A in (9). Thus GO is the angle bisector of EGH. Similarly, HO is the angle bisector of GHE. So their point of intersection, which is O, is the incenter of EGH.Anonymousnoreply@blogger.com