tag:blogger.com,1999:blog-6933544261975483399.post5910893037397911434..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1485: Triangle, Orthocenter, Altitude, Circle, Diameter, Tangent, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-63072607232442926952020-11-27T13:56:51.532-08:002020-11-27T13:56:51.532-08:00https://photos.app.goo.gl/7F46Z7YkTbVG5kkk9https://photos.app.goo.gl/7F46Z7YkTbVG5kkk9c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3140905532061875282020-11-27T08:09:47.355-08:002020-11-27T08:09:47.355-08:00minor correction of the last line:
So HF= 140/HT...minor correction of the last line:<br /><br />So HF= 140/HT=8.75<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77740210514239302872020-11-26T00:40:12.919-08:002020-11-26T00:40:12.919-08:00Draw altitude BP
We have BH . HP = 140
=> TH . ...Draw altitude BP<br />We have BH . HP = 140<br />=> TH . HF = 140 => HF = 8.75c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59255040850781142352020-11-25T16:18:57.398-08:002020-11-25T16:18:57.398-08:00https://photos.app.goo.gl/v7KgdZECAesRBJjr9
Draw a...https://photos.app.goo.gl/v7KgdZECAesRBJjr9<br />Draw altitude BD of triangle ABC ( see sketch)<br />We have B, H, D are collinear ( H is the orthocenter)<br />Quadrilateral ADEB is cyclic so HE.HA=HB.HD= 140<br />In cycle BTDC we have HT.HF=HB.HD=140<br />So HF= 140/HT=8.5<br /><br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com