Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 620.
http://img830.imageshack.us/img830/474/problem620.pngwe have KL//FC and KJ//BIlet FC cut BI at JWe have angle (IEA)= angle (DEF)=35 both angles complement to angle AEDAIBE and ECFD are cyclic quadrilateralsso angle (DCF)= angle (DEF)= 35 ( see picture)and angle (IBA)= angle (AEI)=35so x=180-55-55= 70Peter Tran
Comment(See Problem 620)More generally,Let ∆PQR be the orthocentric triangle of ∆EAD(formed by the feet P,Q,R of the perpendiculars from E,A,D upon the opposite sides AD,DE,AE respectively).Then ∆KJL is similar to ∆PQR with ∠JKL = 180°-2∠AED,∠KJL = 180°-2∠EAD,∠JLK = 180°-2∠ADE.
PravinIt is not clear to me about statement"Then ∆KJL is similar to ∆PQR with∠JKL = 180°-2∠AED, "Please explain.Peter Tran
Let angle AED = y. Then your proof will be as follows:"we have KL//FC and KJ//BIlet FC cut BI at M (You named it as J)We have angle (IEA)= angle (DEF)=90 - y both angles complement to angle AEDAIBE and ECFD are cyclic quadrilateralsso angle (DCF)= angle (DEF)= 90 - y ( see picture)and angle (IBA)= angle (AEI)=90 - ySo angle MBC=angle MCB=yso x=180-y-y= 180-2y" Now it is wellknown that in any triangle ABC,if D,E,F are the feet of the perpendiculars from A,B,C on the opposite sides, and triangle DEF is formed, then its angles are given byD=180-2A,etcIn our problem we have triangle EAD and its orthic triangle is named by me as PQRSo angle QPR = 180-2AED = angle JKL etc