tag:blogger.com,1999:blog-6933544261975483399.post3126282425209138193..comments2022-11-29T01:21:52.908-08:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1059: Triangle, Circumcircle, Orthocenter, Altitude, Perpendicular, Reflection of a Point over a Line, Collinear Points, Simson Line, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-761097836258213922014-11-12T22:45:31.400-08:002014-11-12T22:45:31.400-08:00Firstly, since DM⊥AB, DI⊥BC, DS⊥AC,
thus MIS is t...Firstly, since DM⊥AB, DI⊥BC, DS⊥AC, <br />thus MIS is the Simpson's line of point D. <br /><br />Now since DM=D₃M, DI=D₁I, DS=D₂S, <br />thus D₃D₁D₂ // MIS, with D₂D₃=2×SM. <br /><br />Join DH, and DH cuts MIS at F. <br />We want to show that H lies on D₃D₁D₂. <br />If so, then F is automatically the mid-point of DH and we are done. <br /><br />Let BH cuts AC at J, and cuts the circumcircle at K. <br />Join DK. Let DK cuts AC at L, and cuts MS at N. <br /><br />Since DISC concyclic, thus ∠ISD=∠ICD=∠BCD=∠BKD=∠KDS, <br />so ∠NDS=∠NSD, and so NS=ND. <br />Which means N is the circumcenter of ΔLDS, thus DN=NL. <br />Therefore L lies on the line D₃D₁D₂. <br /><br />Now since HJ=JK, thus ∠HLJ=∠KLJ=∠DLS=∠NSL. <br />So HL//NS which is the Simpson's line MIS. <br />Hence, H lies on D₃D₁D₂ and then F is the mid-point of DH. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-24474631430273861222014-11-12T21:51:15.529-08:002014-11-12T21:51:15.529-08:00http://s25.postimg.org/a4t0mmxmn/pro_1059.png
Let...http://s25.postimg.org/a4t0mmxmn/pro_1059.png<br /><br />Let B’ and C’ are the reflection of H over AC and AB. B’ and C’ will be on circle O.<br />Let HD2 and HD3 cut AC and AB at E and F<br />HDD2B’ and HDD3C’ are isosceles trapezoids and Triangles HB’E and HFC’ are isosceles.<br />∠ (B’HE)+ ∠ (C’HF) = ∠ (HB’E) + ∠ (HC’F)= angle A => ∠ (C’HB’) supplement to ( ∠ (B’HE)+ ∠ (C’HF)) <br />So E, H and F are collinear => D2, H, D3 are collinear<br />M, I and S is located on linear Simpson line so D1, D2 D3 and H are collinear and DF/DH= Di/ D1= 1/2 <br />Obvious D2D3=2. SM<br /> <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com