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
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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]
Why is M_1 the midpoint of E_1F_1?
Deleteconsider problems 781,782,783 and the dilatation with center B and ratio 2.
ReplyDeleteHe renombrado los puntos E y F como P y N.
ReplyDeletec_c = ◯ABC con centro O
A'=S_O(A) ⇒ A' ∈ c_c, ∠A'BA = ∠ACA' = 90°
⇒ A'B ∥ CH, A'C ∥ BH ⇒ [BA'CH] paralelogramo ▱
⇒ ∠A'BH = ∠HCA', M punto medio de HA'
{H,P,B,A'}, {H,A',C,N} concíclicos ◯
⇒ ∠A'PH = ∠A'BH, ∠HNA' = ∠HCA'
⇒ ∠A'PH = ∠HNA'
△HPA' ≡ △HNA' ⇒ PH = HN (qed)
Puede verse el dibujo en https://ilarrosa.github.io/GeoGebra/H_punto_medio.html
Nota: usando el teorema de la mariposa, con la circunferencia de diámetro BC, es inmediato.