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.
Friday, August 22, 2014
Geometry Problem 1038: Isosceles Triangle, Angle, 80, 20, 25, 65 Degrees.
Subscribe to:
Post Comments (Atom)
http://www.mathematica.gr/forum/viewtopic.php?f=22&t=46028
ReplyDeletehttps://apollonius03.files.wordpress.com/2014/09/langley_genel-1.pdf
ReplyDeleteThis may require the reader to perform the construction portion to see the solution.
ReplyDeleteConstruction:
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 – 2alpha
2. From C5 we know angle JHI = angle LEJ, with angle HJI = angle LJE we have
triangle HJI similar to triangle ELJ by AA similarity, therefore IJ is similar to JL
3. 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 = 30
4. 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
ReplyDeleteHere is my solution.
ReplyDeletehttps://www.youtube.com/watch?v=Xg9l71Wayhg
Need to check the steps again. Some of them are not actually proved but assumed
DeleteSolution
ReplyDeleteIn triangle AEC, ∠AEC=180-(CAE+ACE)=180-(80+65=180-145=35. (1)
∠ECB=∠ACB-∠ACE=80-65=15. (2)
Draw DF ∥ CA, F on AB.
Draw ∠DFG =60, G on BC
Suppose point H on AB, so that GH ∥ CA.
So ∠GHA=∠HGC=100 (supplementary to ∠HAC=80 and ∠ACG=80, respectively) (3)
Connect AG and CH, intersect at P
In quadrilateral AHGC, since by construction GH∥CA and ∠HAC = ∠GCA = 80, then AHGC has 2 parallel sides and congruent base angles, so AHGC is isosceles trapezoid.
So legs AH=CG; diagonals AG = CH; AP=CP; HP=GP (diagonal of isosceles trapezoid are equal, and divide each other into segments with lengths that are pairwise equal)
Since AP=CP, then triangle APC isosceles, so ∠CAP=∠ACP=65 (by construction), so ∠APC=180-(65X2)=180-130=50.
So ∠HPG=opposite ∠APC=50
Since HP=GP, then triangle HPG isosceles, so ∠GHP(∠GHC)=∠HGP (HGE)=½(180–50)=½X130=65 (4)
∠AHC=∠GHA-∠GHC. Since from (3) above ∠GHA=100 and from (4) above ∠GHC=65, then ∠AHC= 100-65=35 (5)
From (1) and (5) above, ∠AEC=∠AHC=35, then HC aligns with EC. Since points H and E lie on AB, point H lies on point E.
So GH aligns with GE and GE∥CA. So GEA=65.
So, FEGD is isosceles trapezoid (GE∥DF; base ∠EFD = ∠FAC=80 (correspondent); base ∠FDG=∠ACD=80 (correspondent)
Connect FG intersects ED at O
In isosceles trapezoid FEGE, FG=DE (diagonals) and DO=FO
So triangle FOD equilateral and so ∠FDO(∠FDE)=60.
DF∥CA, so ∠FDG=∠ACG=80; ∠EDG=∠FDG-∠FDE=80-60=20.
In triangle ECD, ∠EDG external, so ∠EDG=∠ECD(ECB)+∠CED
∠CED=∠EDG-∠ECB.
From (2) above ECB=15
So ∠CED=20-15=5 degrees.