## Sunday, November 8, 2020

### Geometry Problem 1482: Right Triangle, Perpendicular, Double Angle, Measurement

Geometry Problem 1482. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below. 1. Draw a)EH perpend AC b)AP bisector of angEAC c)EMT perpend AP
=> Tr DET isosceles, from EM=6 => ET=DE=12

1. https://photos.app.goo.gl/XN6AessMnkA9GsNf6

2. https://photos.app.goo.gl/wrmr2bHVba98ZA3k8

Draw angle bisector AH of angle CAE
Draw EF ⊥AH ( see sketch)
Note that ∠ (CEF)= 2. Alpha and triangle EAF is isosceles
Triangle ABE congruent to AEH and AFH
So EH=HF=BE=6 so EF=12
In triangle EFC , external angle ∠ (DFE)= theta+2. Alpha
In triangle AED, external angle ∠ (EDF)= theta+2. Alpha
So tringle DEF is isosceles and DE=EF=12

3. Let:
AB = a

1.)
3α + Θ +90 = 180 => Θ = 90 - 3α

2.)
Angle x / (2sin(α)) = √(a² + 36)
_________________

From (3) and (4):

x = 12.

4. Extend EB to F such that BE = BF = 6 so that BA bisects < FAE

Then < FEA = 2.Alpha + Theta = 90 - Alpha from which easily < CED = 4.Alpha = < CAF

Hence ADEF is concyclic and since EA bisects < CAF, it follows that DE = EF = 12

Sumith Peiris
Moratuwa
Sri Lanka

5. Fold Tr.ABE along AE such that B' is the reflection of B (ABEB' is cyclic )
m(BAE)=m(EAB')=Alpha=>AB' is angle bisector of m(EAD). Extend EB' to meet AC at F
Since AB' bisects m(EAF) and AB' perpendicular to EF => EF=2.EB'=2.BE=12
m(EFA)=180-m(EFC)=2Alpha+Theta=m(EDF)
=> EDF is an isosceles triangle and ED=EF=12

6. Simple Solution

Extend EB to F such that EB = BF = 6 and AF = FE

Tr.s ABE & ABC are similar and AE bisects < CAF and so
AE/DE = AC / CE = AF / FE = AE / 2.BE

Hence DE = 2.BE = 12

Sumith Peiris
Moratuwa
Sri Lanka