Geometry Problem

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

Click the figure below to see the complete problem 914.

## Saturday, August 24, 2013

### Problem 914: Triangle, Orthocenter, Angle Bisector, Parallel Lines, Midpoint

Labels:
angle bisector,
midpoint,
orthocenter,
parallel,
triangle

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See sketch for location of points O, L, N

ReplyDeleteQuadrilateral ANHL is cyclic with circumcenter O

Since AD is an angle bisector of ∠ (BAC) => ∠ (NAG)=∠(GAL)

And ∠ (NOG)= ∠ (GOL)= 2. ∠ (NAG)=2. ∠ (GAL) => OG is a perpendicular bisector of LN => MN=ML

Let M’ is midpoint of BC.

In right triangle BLC we have M’L=M’B

In right triangle BNC we have M’N=M’B

So M’L=M’N => M’ is on perpendicular bisector of LN.

Both M and M’ on both BC and OG extension => M coincide with M’

Result of this problem show that

1. Perpendicular bisector of NL where N and L are the feet of altitudes of triangle ABC will pass through the midpoint of a side of triangle ABC or

2. Properties of 9 points circle of triangle ABC: perpendicular bisector of NL, where N and L are the feet of altitudes of triangle ABC will pass through another 9 points.

Can you show the sketch, Peter? Thanks.

ReplyDeleteSean N y L los puntos medios de los arcos BC y BAC, respectivamente, del circuncirculo de ABC. Sean J y K las intersecciones de HG y HF con NL, respectivamente.

ReplyDeleteSabemos que las bisectrices AD y AE pasan por N y L, respectivamente. Sea 2a=AH, es conocido el lema AH=2OM, entonces

OM=a,

LJ=NK=2a, (1)

OL=ON=R (circunradio)

MN=R-a,

OJ=R-2a,

MJ=OJ+OM=R-a,

vemos que MN=MJ (2).

Por (1) y (2) M es punto medio de JN y LK que son respectivamente las hipotenusas de los triangulos rectangulos semejantes (por catetos paralelos) JNG y LKF, por lo tanto GM y FM estan sobre una misma recta.

See below for English translation of QueloP's solution.

DeleteIn triangle ABC, if we have AH=2.OM then M will be midpoint of BC. No further proof is required. how do you get AH=2OM ? please explain .

Let N and L be the midpoints of the arcs BC and BAC, respectively, of the circumcircle of ABC. Let J and K be the intersections of HG and HF with NL, respectively.

We know that the bisectors AD and AE pass through N and L, respectively. Let 2a = AH, the lemma AH = 2OM is known, then

OM = a,

LJ = NK = 2a, (1)

OL = ON = R (circumrad)

MN = R-a,

OJ = R-2a,

MJ = OJ + OM = R-a,

We see that MN = MJ (2).

By (1) and (2) M is the midpoint of JN and LK which are respectively the hypotenuses of triangles similar rectangles (by parallel hinges) JNG and LKF, therefore GM and FM are on the same line.

Begin by drawing the orthic triangle within triangle ABC. Name this triangle PLW with P being the bottom vertex, L being the left vertex, and W being the right vertex. The orthic triangle is the triangle created by connecting the points on triangle ABC that are created by the altitudes intersecting the sides. Then draw the circumcircle of the orthic triangle. The circumcircle is the circle that has its center at the circumcenter of the orthic triangle. It has been proven that the circumcircle of an orthic triangle contains the midpoints of the sides of the original triangle. From here, you must find the perpendicular bisectors of the orthic triangle to fin the center of the circumcircle. It appears that segment FG is a perpendicular bisector to the PL. If you can prove this to be true, then that will prove that M is the midpoint of line segment BC because it will show that there is a radius that goes to point M, and the circumcircle includes the midpoints of the sides. This can be proved using triangle PFL. Line segment FG intersects triangle PFL through point F. If we can prove that triangle PFL is an isoscles triangle, then segment FG will cut triangle PFL into two congruent right triangles, meaning that FG is a perpendicular bisector to PL. To prove this, draw a circle with a center at point F. Line segments FH and FP are both radii to circle F, therefore, they are congruent. This makes triangle PFL isoscles, and FG a perpendicular bisector to PL. Because of this, M must be the midpoint of BC

ReplyDeleteConsidering usual triangle notations, let E be the foot of the perpendicular from B to AC

ReplyDeleteO be the center of the rectangle AFHG

P be the intersection of AB and FG

Q be the intersection of BE and FG (FPOQGM are collinear)

Observe that AFHGE are concyclic points

we know m(EAF)=90+A/2 => m(FOE)=180-A (Angle at center is twice at the circumference)

Since m(PAE)=A and m(POE)=180-A => POEA are concyclic ------(1)

A bit of angle chasing leads us to the following

m(FGA)=90-(C+A/2)

m(BDA)=180-(B+A/2)

=> m(DMG)=m(BMG)=90+(C-B) -------------(2)

In the triangle BQM, m(BQM)=180-m(QBM)-m(BMQ)=180-(90-C)-(90+C-B)=B ----------(3)

From (1) & (3), we can observe that m(EOQ)=A,m(OQE)=B=>m(OEQ)=C ------(4)

From (1) & (4), m(PEQ)=m(PEO)+m(OEQ)=m(PAO)+C=90-B+C--------------(5)

From (2) & (5), observe that m(BMG)=m(BMP)=90+(C-B)=m(PEQ)=m(PEB)

Hence BPEM are concyclic and m(BEM)=m(BPM) ---------(6)

m(BPM) can be derived by considering the triangle BPM and is equal to 90-C

Hence m(BEM)=90-C and m(MEC)=C => BEM and EMC are isoscelese triangles with BM=ME=MC => M is circumcenter of BEC and the mid-point of BC