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Level: Mathematics Education, High School, Honors Geometry, College.

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## Sunday, October 8, 2017

### Geometry Problem 1349: Three Squares, 90 Degrees, Perpendicular Lines, Diagonal. Math Infographic

Labels:
diagonal,
geometry problem,
perpendicular,
three squares

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Problem 1349

ReplyDeleteLet < AHB = < HBG = < AGB = θ.

AC2 = 2.CE2 = CE.CG, hence < CAE = < AGB = θ.

Since < CBP = < CAP =θ,

ABCP is concyclic and so

< CPE = < ABC = 90

Sumith Peiris

Moratuwa

Sri Lanka

Interestingly also AE = 5.CP

DeleteProblem 1349

ReplyDeleteIn the extension of AB to B above point K such that AB=BK=>triangleKAD=triangleAFR=E.So

KD=AE and KD perpendicular to AE.If the AE intersects the KD in Q and the BC in M. Then

BM=MC and EQ/QA=EM/AD=3/2 =>EQ/EA=3/5 (1).Is EP/PA=BE/AH=2/3 =>EP/EA=2/5(2).

From (1) and (2)=>EP/EQ=2/3=>EP/PQ=2/1=EC/CM=>CP//MQ=>CP is perpendicular in AE.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

https://photos.app.goo.gl/vydebtzBccTnATBf1

ReplyDeleteLet BH meet EF at K

Let AB=BC=CD=AD= 1

So EK= 2/3 and AE= sqrt(5)

Triangle ABP similar to EKP ( case AA)

PE/PA= EK/AB= 2/3

So PE=2/sqrt(5) and PA=3/sqrt(5)

And PE/EB= 1/sqrt(5) and CE/AE= 1/sqrt(5)

Triangle ECP similar to EBA ( case SAS)

So ∠ (CPE)= ∠ (EBA)= 90 degrees