Geometry Problem

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

Click the figure below to see the complete problem 784.

## Monday, July 16, 2012

### Problem 784: Triangle, Orthocenter, Altitude, Midpoint of a side, Perpendicular

Labels:
altitude,
midpoint,
orthocenter,
perpendicular,
side,
triangle

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http://img402.imageshack.us/img402/4156/problem784.png

ReplyDeleteDraw lines per attached sketch

Note that P and Q are midpoints of BC’ and CB’

Quadrilaterals MHEP and MFQH are cyclic

In quadrilateral MHEP, ∠MEH=∠MPH=∠PHC’

In quadrilateral MFQH , ∠MFH=∠MQH=∠QHB’

∆BHC’ similar to ∆ CHB’ => HC’/HB’=BC’/CB’=(.5.BC’)/(.5.CB’)=PC’/QB’

So ∆PHC’ similar to ∆QHB’ …( case SAS)

And ∠PHC’=∠QHB’

∠MEH=∠MFH => triangle MFE is isosceles => H is the midpoint of EF

Let O be the circumcenter of triangle ABC. Also, let M₁ be the midpoint of AH. From M₁ build E₁F₁ perpendicular to OM₁

ReplyDeleteOMHM₁ is a Parallelogram

OM₁ ǁ MH → E₁F₁ ǁ EF

M₁→ midpoint [E₁F₁] →H midpoint of [EF]

consider problems 781,782,783 and the dilatation with center B and ratio 2.

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