## Friday, November 4, 2016

### Geometry Problem 1283 Two Equilateral Triangle, Perpendicular, Midpoint

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1283.

1. https://goo.gl/photos/96rSC1H87PN4Xb7g7

Let N is the projection of M over CDE
In triangle EMD, MN is the median and altitude from M => MED is isosceles triangle
Connect CM , CM ⊥AB and BE⊥EC => quadri. CMBE is cyclic
in cyclic quadrilateral CMBE ∠MDC=∠MBC=60 degrees => Triangle EMC is equilateral

2. Problem 1283
Suppose that MN is perpendicular to CE ( DN =NE ,ABED is trapezoid ) so triangle MED is isosceles (ME=MD).But <AMC=90=<ADC then AMDC is cyclic. So <MDA=<MCA=60/2=30.
Then <MDE=90-30=60.Therefore triangle MDE is equilateral.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

3. BECM is cyclic with BC as diameter so
< MEC = < MBC = 60

ACDM is cyclic with AC as diameter so
< MDE < MAC = 60

So 2 angles of Tr. MDE = 60 and is hence equilateral

Sumith Peiris
Moratuwa
Sri Lanka