tag:blogger.com,1999:blog-6933544261975483399.post7208263660123589962..comments2024-04-22T04:55:16.794-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1293 Triangle, Altitudes, Two Squares, Center, Concurrent Lines, MidpointAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-46246667089037061622016-12-08T11:06:16.251-08:002016-12-08T11:06:16.251-08:00m(BHA) = m(BOA) = 90 => OABH is cyclic quadrila...m(BHA) = m(BOA) = 90 => OABH is cyclic quadrilateral<br />since m(BAO) = 45 => m(BHO) = 45 => m(OHA) = 45 --------(1)<br />Extend AH to meet BF at P and consider the triangle BAP<br />we have <br />m(ABP) = B+45<br />m(BAP) = 90-B<br />=> m(BPA) = 45 ----(2)<br />From (1) and (2), we can say OH||BF<br />Since O is mid-point of BE, from mid-point theorem, M is the midpoint of EF<br />similarly QCJB can be found cyclic and thus JQ intersects EF at MAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45953053247561621322016-12-08T10:06:29.642-08:002016-12-08T10:06:29.642-08:00Quad. AOBH is cyclic,Angle BOH= Angle BAH =90-B,An...Quad. AOBH is cyclic,Angle BOH= Angle BAH =90-B,Angle OBQ=90+B,Hence OH is parallel to BF. Since O is midpoint of BE, line OH will pass pass through midpoint of EF.<br />Similarly QJ is parallel to BE and passes through midpoint of EF. Hence QJ, OH and EF are concurrent at mid point of EF.Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-81265782827507452432016-12-08T09:27:55.702-08:002016-12-08T09:27:55.702-08:00Connect OA and OB
Observe that qua. OBHA is cyclic...Connect OA and OB<br />Observe that qua. OBHA is cyclic<br />So ∠BAH=∠BOH<br />In triangle ABH , ∠BAH complement to ∠B => ∠BOH complement to ∠B<br />So ∠BOH supplement to ∠OBF => OH //BF<br />Since O is the midpoint of EB and OH//BF=> OH cut EF at midpoint M<br />Similarly QJ also cut EF at midpoint M<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com