Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view the complete problem 1038.
This may require the reader to perform the construction portion to see the solution.Construction:C1. Reflect ED across EC yielding ED’.C2. Reflect ED’ across EA, label D” as H. Label the intersection of D’H as G. Note that by construction EG is both an altitude and angle bisector.C3. Construct D’J perpendicular to EH, label the intersection with EG as I. Note that since EG and D’J are both altitudes, I is the orthocentre.C4. Construct HK through I, from 4 we know HK is perpendicular to ED’.C5. Construct EL parallel to HK such that L is collinear with D’J.Proof:1. Let angle JEI = alpha = angle KEI. So we have:angle JIE = angle KIE = angle HIG = 90 – alpha , therefore angle JHI = 90 – 2alpha2. From C5 we know angle JHI = angle LEJ, with angle HJI = angle LJE we havetriangle HJI similar to triangle ELJ by AA similarity, therefore IJ is similar to JL3. EJ = EJ; angle EJI = angle EJL; and IJ ~ JL, therefore triangle ELJ is similar to triangle EIJ by SAS similarity. Therefore angle LEG = angle IEJ, 90 – 2aplha = alpha, therefore alpha = 304. From the givens of the problem we have angle AEC = 35, therefore angle GED’ = 35 – x = angle JEI (HEG), so we have 35 – x = 30, therefore x = 5. Q.E.D.
I do not understand stage 3 of the proof: ELHI is a trapezoid with perpendicular diagonals (not a parallelogram)you mix equal and similar