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

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

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## Monday, November 7, 2016

### Geometry Problem 1284 Two Equilateral Triangle, Midpoint, Congruence

Labels:
congruence,
equilateral,
midpoint,
triangle

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https://goo.gl/photos/49eNJ5Yuiq4nrZ5P8

ReplyDeleteDraw M on BF such that FM=FB ( see sketch)

We have triangle ACM is equilateral

Triangle ACE congruent to MCD ( case SAS)

And triangle MCD is the image of ACE in the rotational transformation center at C, rot. Angle= 60 degrees.

So AE=MD and angle(AE, MD)= 60

And FH=1/2.AE and FH//AE

FG=1/2.MD and FG//MD

So FG=GH and angle(FG, FH)= angle (AE,MD)= 60 degrees

And FGH is equilateral

Problem 1284

ReplyDeleteLet the equilateral triangle AEP (PA,PE intersects the BD, not the extensions ). Then triangle ABP=triangle ACE (AB=AC,AP=AE, <BAP=60-<PAC=<CAE).So BP=CE=CD,but

<PBC+<BCD=<ABP-60+<BCD=<ACE-60+<BCD=240-60=180, then BP//CD so the BPDC is

Parallelogram .Then the points P,G and C are collinear with PG=GC.So GF=AP/2=AE/2=FH

and <HFG=<EAP=60 (GF//AP,FH//AE).Therefore triangle FGH is equilateral.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

Complete the rhombus ABCX.

ReplyDeleteTriangles ACE ≡ XCD, SAS.

So AE = XD and hence FH = FG from the mid point theorem in triangles ACE

and XBD.

Similarly we can show that GH = FH by completing rhombus CDEY.

Hence FG = FH = GH

Sumith Peiris

Moratuwa

Dear Antonio

ReplyDeleteMy solution does not appear yet, hence resenting.

Complete the rhombus ABCX.

Triangles ACE ≡ XCD, SAS.

So AE = XD and hence FH = FG from the mid point theorem in triangles ACE

and XBD.

Similarly we can show that GH = FH by completing rhombus CDEY.

Hence FG = FH = GH

Sumith Peiris

Moratuwa

Sri Lanka