## Sunday, October 8, 2017

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

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

Details: Click on the figure below. 1. Problem 1349
Let < 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

1. Interestingly also AE = 5.CP

2. Problem 1349
In 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

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

Let BH meet EF at K
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

4. Prove by similar triangles: △CPE similar to △ABE

Let CE=1. Then BE=2, AH=3
△BPE similar to △HPA (AAA) so AP:PE=3:2
AE=SQRT(5)
PE=2(SQRT(5)) / 5

Since <AEB=<CEP (common angle)
AE/BE= SQRT(5)/2
CE/PE= 1/ 2(SQRT(5)) / 5 = 5 / 2(SQRT(5)) = SQRT(5) / 2
Hence △CPE similar to △ABE ( ratio of 2 sides, inc.<)
Since <ABC=90 degrees, <CPE is therefore 90 degrees. (proved)

5. Problem !349 is more amenable for analytic proof:
Let A = (0,0), D = (1,0), B = (0,1)
Then C = (1,1), E = (2,1), F = (2,0), H = (3,0)
Equations: BH: x + 3y = 3, AE: x - 2y = 0,
their intersection P = (6/5,3/5)
Slopes: of AE = 1/2, of CP = -2
Hence their product being -1, they are perpendicular.

6. Tyler Owen Eli ArianeJanuary 19, 2018 at 8:32 AM

Great work! you guys really know what you are doing! WOW just WOW!