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

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

Details: Click on the figure below.

## Saturday, December 3, 2016

### Geometry Problem 1291 Three Squares, Collinear Vertices, Midpoint, Congruence

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https://goo.gl/photos/mimSHwEKYejTtxPm9

ReplyDeleteLet M,N, P, O are the projection of J, E, G and B over AC

Note that triangles CMJ congruent to ENC . Triangle ENA congruent to APG

So CM=EN= u and JM=CN=v ( see sketch)

and EN=AP=u and AN=PG= v1

OP=OA+AP=w+u =OC+CM=OM => O is the midpoint of PM

AC=AN+NC=v1+v=BD=2.w=> w=1/2(v+v1)

In trapezoid GPMJ , middle base OB’=1/2(PG+MJ)= ½(v1+v)=OB

So B coincide to B’ => G, B, J are collinear and B is the midpoint of GJ

Prolem 1291

ReplyDeleteTriangle GAB=triangle AED (AG=AE,AB=AD,<GAB=90+<EAB=<EAD), so GB=ED and <GBA=<EDA.But triangle ECD=triangle JCB (EL=LJ,CD=CB, <ECD=<BCJ=90+<ECB ), so

BJ=DE and <CBJ=<CDE.Then GB=BJ and <GBJ=<GBA+<ABC+<CBJ=90+<ADE+<EDC=90+90=180. Therefore the points G,B and J are

collinear.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

It is also true that in this question FB=BH and FB is perpendicular to BH. Any suggestions for proof?

ReplyDeleteSuggestions: See the problem 496 at

Deletehttp://www.gogeometry.com/problem/p496_triangle_two_square_side_concurrent_90_degrees.htm

Found it: Let P be the midpoint of GE; Q the midpoint of EJ

ReplyDeletePEQB is a parallelogram (PB is midpoint segment in Triangle EGJ)

Triangle FPB congruent with Triangle BQH

Hence, FB=BH

<FBP+<FBH+<HBQ+<BQE=180 (cointerior angles in p'gram) I

<HBQ+<BQE+<QHB=90 (adds up to 180 in triangle) II

<FBP=<QHB (matching angles in congruent triangles)

Subtract II from I: <FBH=90