Sunday, April 9, 2017

Geometry Problem 1328: Two Squares Side by Side, Perpendicular, 90 Degrees, Congruence

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 1328
Is AE perpendicular in CG and HK ,then CG//=HK.But <CHG=<HGD=45=<CAG,so
A,H,C and G are concyclic (AG//HC) .So AHCG is isosceles trapezoid , then
AH=CG. Therefoore AH=HK.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

2. AC bisector of HAE => HAD=45°+HAE, AGC=45°+EGC
=> HAG=AGC (CAE=EGC)

3. Triangle HAE is symmetrical with respect to AC
==> Triangle AKH isósceles in H ===> AH = HK

4. https://goo.gl/photos/nFqf1YZ178sWAnWn6

Draw CG and HM ( see sketch)
Per the result of problem 1327 AE⊥CG => HK//CG
So HCGK is a parallelogram => HK= CG
triangle HCE is 45-45-90 triangle so CH= CE
HB= CB-CH= CD-CE= ED=DG
Triangle HAM congruent to CDG ( case SAS)
So ∠ (HAM)= ∠ (CGD)= ∠ (HKM) => AHK is isoceles
So HA=HK

5. Reference my proof for Problem 1327 AE and CG are perpendicular as are AE and HK

Hence HK//CG and so HCGK is a parallelogram.

So HK = CG

Now AH = AE since Tr.s AHC and AEC are congruent SAS.

But AE = CG since Tr.s AED and CGD are congruent SAS.

So HK = CG = AE = AH

Sumith Peiris
Moratuwa
Sri Lanka

6. Without using the results of problem - 1327
Drop a perpendicular from H to AD and denote it as M.
Since m(AJH) = m(AMH) = m(ABH) = 90 => ABHJM are concyclic
Therefore m(MJH) = m(MAJ) and since HM = AD => triangles AED congruent to HKM (ASA)
=> AE = HK --------- (1)
Since m(HEC) = 45 => HCE is right angle isosceles and AC bisects HE => AHE is isosceles triangle
=> AH = AE ---------- (2)
From (1) and (2) => AH = HK