tag:blogger.com,1999:blog-6933544261975483399.post1578763887101999594..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 400. Triangle, Angle bisector, Circumcircle, Perpendicular, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-6933544261975483399.post-84845348847486959182023-12-02T20:25:18.822-08:002023-12-02T20:25:18.822-08:00Let <DCE=<DCB=x
<OCE+<OFC=180
OECF is ...Let <DCE=<DCB=x<br /><OCE+<OFC=180<br />OECF is a cyclic quad<br /><HOG=<OCE=2x (ext. < cyclic quad)<br /><br />Consider triangle CFH<br /><OHG=<FHC=90-x (< sum of triangle)<br /><OGH=180-<OHG-<HOG=90-x=<OHG<br />So, OH=OG (sides opp. eq. <s)<br /><br />Consider triangle OGD & triangle OHC<br />OD=OC (radii)<br /><ODH=<ODC=<OCD=<OCG (base <s, isos triangle)<br /><OGD=<OGH=<OHG=<OHC (Proved)<br />Triangle OGD congruent to triangle OHC (AAS)<br />So, DG=HC (corr. sides of congruent triangle)<br />So, DH+HG=CG+GH<br />DH=CGMarcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-61961326089411382582016-10-21T16:52:18.441-07:002016-10-21T16:52:18.441-07:00More or less the same as the last few but with a f...More or less the same as the last few but with a few extra steps included<br />1. Right triangle CEG is similar to right triangle CFH because of the angle bisector.<br />2. So angle CGE = angle HGO = angle FHC and triangle OHG is isosceles.<br />3. OD = OC since they are radii.<br />4. So CDO is isoscleses and angle ODC = OCD.<br />5. Angle OHD = OGC since they are supplementary to 2 congruent angles.<br />6. We now have 2 out of 3 angles and 2 out of 3 sides congruent in ODH and OGC which is more than enough to infer a missing angle and use either SAS or ASA to show the triangles are congruent.<br />7. So DH = CG.<br /><br />Benjamin Leishttps://www.blogger.com/profile/10974191081762367425noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86411199585932565062016-10-19T09:47:20.306-07:002016-10-19T09:47:20.306-07:00OGH isosceles ==>
perpendicular from O to CD bi... OGH isosceles ==><br />perpendicular from O to CD bisect GH, and also bisect CD ==> <br />GC = DHIgnacio Larrosa Cañestrohttps://www.blogger.com/profile/03014700623279846626noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74483555726942195612016-06-17T04:53:57.950-07:002016-06-17T04:53:57.950-07:00Triangles HFC and EGC are similar
So OH = OG
So...Triangles HFC and EGC are similar <br /><br />So OH = OG<br /><br />So Tr.s ODH and OGC are congruent ASA and the result follows<br /><br />Sumith Peiris<br />Moratuwa<br />Sri Lanka Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-43914201592497651812009-12-10T01:52:08.453-08:002009-12-10T01:52:08.453-08:00step 1 and 2 need third condition of congruence,
a...step 1 and 2 need third condition of congruence,<br />angle DOH = angle COG ? verify please<br /><br />my solution<br />to draw OK perpendicular to HG give <br />1) OK is median => HK = KG<br />2) OK is diameter perpendicular to chord DC => DK=KC<br />so DH=DK-HK<br /> and GC=KC-KGc .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-46874496300605204612009-12-09T15:36:24.623-08:002009-12-09T15:36:24.623-08:00W/o having to refer to any other figure or any con...W/o having to refer to any other figure or any construction, we can easily see that quad. OFCE is concyclic and hence ang. HOG = ang. C while ang. OGH = ang. EGC = 90-C/2. Therefore, ang. OHG = 180 – C - (90-C/2)= 90 – C/2 or ang. OHG = ang. OGH. Hence etc.<br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18244071379688261902009-12-09T14:21:12.729-08:002009-12-09T14:21:12.729-08:00Thanks!Thanks!Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39976690826124349502009-12-09T13:30:19.149-08:002009-12-09T13:30:19.149-08:00reason for step 1:
http://i49.tinypic.com/69omdj.g...reason for step 1:<br />http://i49.tinypic.com/69omdj.gif<br /><br />and<br />step 1 and step 2 are enough for step 3, i thinkAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-71814254967004326362009-12-09T06:39:36.029-08:002009-12-09T06:39:36.029-08:00suggest to others
1) draw HP perpendicular to AC ...suggest to others<br /><br />1) draw HP perpendicular to AC ( P on AC )<br />2) draw OK perpendicular to HG ( K on HG )<br /><br />about first comment: step 1 and step 2 are not enough for step 3c .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90875555308683643952009-12-09T06:13:13.417-08:002009-12-09T06:13:13.417-08:00Reasons for step 1: |OG|=|OH| ?Reasons for step 1: |OG|=|OH| ?Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-1525203597099500842009-12-09T01:41:43.701-08:002009-12-09T01:41:43.701-08:00|OG|=|OH|
|OC|=|OD|=r
|GC|=|DH||OG|=|OH|<br />|OC|=|OD|=r<br />|GC|=|DH|Anonymousnoreply@blogger.com