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
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
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ReplyDeleteLet 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
Problem 1283
ReplyDeleteSuppose 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
BECM is cyclic with BC as diameter so
ReplyDelete< 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