Geometry Problem

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

Click the figure below to see the complete problem 785.

## Wednesday, July 18, 2012

### Problem 785: Equilateral Triangle, Parallel to a side, Perpendicular to a side

Labels:
equilateral,
parallel,
perpendicular,
triangle

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Let K be the mid-point of AC, such that BK is the altitude.

ReplyDeleteNote that triangles ADE and EFC are both equilateral,

and also DBFE is a parallelogram, so

AE = DE = BF

EC = EF = DB

Thus

a + d + f

= d + EC + AE + f

= d + DB + BF + f

= GB + BH

= 2*MK + 2*KN

= 2*(MK + KN)

= 2x

Let AE = b, then AD = b, EC = a-b, FC = a-b

ReplyDeleteThen

AM = ADcos60 = (b-d)/2

NC = CFcos60 = (a-b-f)/2

by considering the length of AC = AM + MN + NC,

a = (b-d)/2 + x + (a-b-f)/2

x = (a+d+f)/2

q.e.d.

Triangles ADE and CFE are equilateral

ReplyDeleteSo DE + EF = AE + EC = a

Draw DJ, FK each perpendicular to AC

x = ME + EN = (MJ + JE)+(EK + KN)

= [(GD + DE) + (EF + FH)]cos 60°

= [GD + (DE + EF) + FH]/2

= (d + a + f)/2