## Thursday, November 10, 2016

### Geometry Problem 1286 Triangle, Quadrilateral, 60, 135, 105 Degree Angles, Congruence, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click on the figure below to view more details of problem 1286. 1. Problem 1286
Let AB=a, BC=b, CD=c,AD=d. The sides AB and CD intersects at point K then triangle KAD is equilateral (<ADC=360-60-135-105=60). Draw BE//AD( the point E lies on the CD).Then ABED is isosceles trapezoid (AB=ED).So AC=BD=AE or triamgle ACE is isosceles .Then <AED=<ACK and <DAE=<KAC.But triangles KAC=DAE and AB=ED=KC, KB=CD. Draw CF
perpendicular in KB, then KF=KC/2=a/2,CF=√(3 )a/2=BF(<KCF=30,<FCB=<FBC=45).But b^2=2CF^2 or b/a=√(6 )/2.
c=CD=KB=KF+FB=a/2+√(3 )a/2 =a(√(3 ) +1)/2 or c/a=(√(3 ) +1)/2.
And d=AD=KA=AB+BK=a+c=a+ a(√(3 ) +1)/2=a (√(3 ) +3)/2 or d/a=(√(3 ) +3)/2 .
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

2. https://goo.gl/photos/YmHy2K7uN38Yp5xY8
Let AB meet CD at E
Draw BF//AD , AH⊥CD, CG⊥BE ( see sketch)
We have AED and BEF are 60-60-60 triangles
AHD , AHE and CGE are 30-60-90 triangles
And CCG is 45-45-90 triangle
Observe that H is the midpoint of CF and DE ;and AF=BD=AC
Let u= CG
So BG= u and BC= u.sqrt(2)
EC=2u/sqrt(3)
EG= u/sqrt(3)
EF=BE=u(1+1/sqrt(3))
CF= EF-EC=u(1-1/sqrt(3))
AB=FD=CE=2u/sqrt(3)
CD=EF=u(1+1/sqrt(3))