tag:blogger.com,1999:blog-6933544261975483399.post1131069788527513006..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 779: Triangle, Altitude, Orthocenter, Vertex, Midpoint, Side, Angle, 90 DegreesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-27935790054532684532012-07-09T01:14:14.708-07:002012-07-09T01:14:14.708-07:00∆ADC is rt angled at D, and F is the midpoint of t...∆ADC is rt angled at D, and F is the midpoint of the hypotenuse AC.<br />So AF = DF, ∠ADF = A.<br />∆BDH is rt angled at D, and M is the midpoint of the hypotenuse BH.<br />So ∠BDM = ∠DBM = 90° - A (note BH ⊥ AC)<br />Follows ∠ADF + ∠BDM = 90°<br />Hence ∠FDM = 90°Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-14893657367898569212012-07-08T08:43:16.735-07:002012-07-08T08:43:16.735-07:00http://img37.imageshack.us/img37/6047/problem779.p...http://img37.imageshack.us/img37/6047/problem779.png<br /><br />Draw nine-points circle per attached sketch<br />This circle will pass through feet of altitudes, midpoint of each side and midpoint from orthocenter to each vertex.<br />Since ∠ (MNF)=90 => MF is a diameter<br />So ∠ (MDF)=90Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65496790558650568612012-07-08T07:43:22.229-07:002012-07-08T07:43:22.229-07:00Triangle ADC is similar to triangle DHB
And since...Triangle ADC is similar to triangle DHB <br />And since F is the mid point of AC and M is the mid point of BH, so triangle ADF is similar to DHM and DFC is similar to DHM<br />So angle FDM is same as angle ADC , which is 90<br />QEDW fungnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83327152721392280442012-07-08T07:31:48.006-07:002012-07-08T07:31:48.006-07:00By properties of orthocenter, and cyclic quadrilat...By properties of orthocenter, and cyclic quadrilateral, <br />Angle DAC = Angle DHB<br /><br />Thus, <br />Triangle DAC ~ Triangle DHB<br /><br />Now, consider a rotation 90 degrees clockwise about point D, <br />DB maps to a line segment on DC, <br />DH maps to a line segment on DA. <br /><br />So, after the above rotation, <br />triangle DHB will become the same position as triangle DAC, only with different size. <br /><br />Hence, DM will map to a line segment on DF. <br />Which means, angle FDM = 90 degrees.Jacob Hanoreply@blogger.com