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

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

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## Tuesday, August 16, 2016

### Geometry Problem 1246: Triangle, Orthocenter, Altitude, Midpoint, Perpendicular

Labels:
altitude,
midpoint,
orthocenter,
perpendicular,
triangle

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Problem 1246

ReplyDeleteLet AH1,CH1,EH2 and DH2 intersects the AB,CB at K,L,M and N respectively then <ALC=90=<AKC so ALKC is cyclic. Then AH1.H1K=LH1.H1C. But AH1.H1K is power the point H1 in the circle ( A,E,K)

with center M1 and LH1.H1C is power the point H1 in the circle (D,L,C) with center M2.

Similar in triangle ADE the point H2 it has an equal forces as to their circles (D,N,C) and (A,E,M) with center M2, M1 respectively .So H1H2 is radical axis of cycles with center Μ1,Μ2.Τherefore H1H2 is perpendicular at M1M2.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

Excellent solution Apostolis

DeleteSee link below for the sketch of Apostolis's solution

https://goo.gl/photos/vsQkRbcfLejMHaTP8

Peter Tran

Thank you Peter for your kind words and for the figure, but I do not know thee

Deletethe way to move the shape.

apostolis manoloudis