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

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## Monday, April 10, 2017

### Geometry Problem 1329: Two Squares Side by Side, Parallel, Perpendicular, 90 Degrees, Congruence

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90,
congruence,
degree,
geometry problem,
parallel,
perpendicular,
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Problem 1329

ReplyDeleteIs CM//EG then ECMG=parallelogram.So CE=MG.But <EHC=<CEH=45=<EGM then

HC=CE=MG so HCMG is isosceles trapezoid (cyclic) or HM=CG.But the AHCG is isosceles

Trapezoid (cyclic) and so the points A,H,C,M and G are concyclic.Now <HAM=<HGM=45

or arc AH=arc HM or AH=HM and <AHM=<AGM=90.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

Problem 1329 Solution 2

ReplyDeleteSuppose that the BC intersects GM in point N then NG=NH=AB (<GHN=45=<HGN=<MCN)

and CN=NM=DG=BH(AG=BN or AD+DG=BH+HN or DG=BH).Istriangle ABH=triangle HNM

so AH=HM and <BAH=<MHN but <BAH+<BHA=90 or <BHA+<MHN=90.Therefore

<AHM=90.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

https://goo.gl/photos/rZBQU3ty6Rsrpccq6

ReplyDeleteBC meet FG at N

Observe that HCE and CNM are 45-45-90 triangles

So HB=ED=CN=DG

Triangles ABH and HNM are congruent => AH=HM and angle AHM= 90

Since HCE is right-angled isosceles, => BH=ED

ReplyDeleteAlso CM||EG => MG=CE => HCMG is an isosceles trepezoid

Join CG and let m(GCD)=x => m(GCM) = 45-x = m(GHM) -----------(1)

Drop a perpendicular from H to meet AG at P and since HPG is isosceles triangle => m(PHG) = 45 --------(2)

Also the triangles ABH and CDG are congruent (SAS) => m(BHA) = 90-x => m(AHP) = x -------------(3)

Therefore m(AHM) = m(AHP)+m(PHG)+m(GHM) = 90

Since m(AGM) = m(AHM) = 90 => AHMG are concyclic

Join AM and since m(MGH) = 45 => m(MAH) = 45 and the triangle AHM is right-angled isosceles => AH=HM

To prove AH=HM:

ReplyDeleteLet intersection point of CM and EF be point I.

<HGM+<GHM = <HMC+<CMF (ext.<)

Since <HGM=45° and <HMC=<GHM

<CMF=45°

By using this result, it can be deduced that triangle HCE,CEI and IFM are isosceles.

Let length of side of large square be 1.

BH^2+1^2=AH^2

(HC+EF)^2+(CE+FM)^2=HM^2 (both Pyth.)

BH+HC=CE+ED=1

HC=CE (isos.)

So BH=ED

HC+ED=1

Since ED=EF,

HC+EF=1

CE=EI (isos.)

IF=FM (isos.)

EI+IF=EF

Hence CE+FM=1

ED^2+1^2=AH^2

1^2+ ED^2=HM^2 (both identical)

Hence AH=HM

To solve <AHM=90°,

ReplyDeleteconstruct point J so that CDGJ is a rectangle and BC extended meets J.

Triangle ABH is congruent to HMJ

<BAH=<JHM

<BAH+<BHA+<90°=180°

<JHM+<BHA+<AHM=180°

Hence <AHM=90°

Let the larger square be of side a and the smaller square be of side b

ReplyDeleteComplete Rectangle CEFN

Triangle HNG is Right Isosceles so HN = a

Triangle CNM is Right Isosceles so NM = b

Further BH = (a+b) - a = b

So Triangles AGH & GNH are congruent SAS (AB = HN = a, BH = NM = b & the included angle is 90)

Hence HM = AH

Also < AHM = 90 since < AHG & < MHN (= < GAH) are complementary

Sumith Peiris

Moratuwa

Sri Lanka