Geometry Problem. Post your solution in the comment box below.

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

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## Sunday, April 9, 2017

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

Labels:
congruence,
geometry problem,
perpendicular,
square

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Problem 1328

ReplyDeleteIs 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

AC bisector of HAE => HAD=45°+HAE, AGC=45°+EGC

ReplyDelete=> HAG=AGC (CAE=EGC)

Triangle HAE is symmetrical with respect to AC

ReplyDelete==> Triangle AKH isósceles in H ===> AH = HK

https://goo.gl/photos/nFqf1YZ178sWAnWn6

ReplyDeleteDraw 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

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

ReplyDeleteHence 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

Without using the results of problem - 1327

ReplyDeleteDrop 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