## Monday, April 10, 2017

### Geometry Problem 1329: Two Squares Side by Side, Parallel, 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 1329
Is 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

2. Problem 1329 Solution 2
Suppose 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

3. https://goo.gl/photos/rZBQU3ty6Rsrpccq6

BC 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

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

5. To prove AH=HM:
Let 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

6. To solve <AHM=90°,
construct 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°