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.

## Sunday, November 8, 2020

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

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Draw a)EH perpend AC b)AP bisector of angEAC c)EMT perpend AP

ReplyDelete=> Tr DET isosceles, from EM=6 => ET=DE=12

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

Deletehttps://photos.app.goo.gl/wrmr2bHVba98ZA3k8

ReplyDeleteDraw 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

Let:

ReplyDeleteAB = a

1.)

3α + Θ +90 = 180 => Θ = 90 - 3α

2.)

Angle x / (2sin(α)) = √(a² + 36)

_________________

From (3) and (4):

x = 12.

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

ReplyDeleteThen < 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

Fold Tr.ABE along AE such that B' is the reflection of B (ABEB' is cyclic )

ReplyDeletem(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