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#1 2015-08-20 15:12:06

phanthanhtom
Member
Registered: 2012-06-22
Posts: 290

A triangle has concurrent median, altitude and angle bisector

In triangle ABC, the median AD, altitude BH and angle bisector CE are concurrent. Prove that (BA + CA)(BC^2 + CA^2 - AB^2) = 2*BC*CA^2.

Please provide detailed proof.

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#2 2015-08-21 06:28:07

Bob
Administrator
Registered: 2010-06-20
Posts: 8,914

Re: A triangle has concurrent median, altitude and angle bisector

hi phanthanhtom

I've tried this, on and off, all day and not got there yet.

I have got some ideas, which I'll share, as it might help you, or someone else to finish the job.

54zdEC8.gif

Because D is the midpoint of BC and BHC is 90, a circle, centred on D, will go through B, C  and H.

Using the cosine rule

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2015-08-23 02:45:57

Bob
Administrator
Registered: 2010-06-20
Posts: 8,914

Re: A triangle has concurrent median, altitude and angle bisector

hi phanthanhtom

OK.  Still no fresh progress so I thought I'd construct a triangle with these properties and take some measurements.  (note: If I click a line and measure it I get this format eg. mBA (overlined) = whatever.  If I select the end points I get this eg.  CA = whatever.  There is no other significance to the occasional 'm'.)

ruDLAhm.gif

To construct this, I first drew BC and found the midpoint.  I made that the centre of a circle with radius DC and chose a point H.

I bisected angle ACB and marked point G where BH and CE intersect.

Then I extended DG and CH to find point A.

Thus, AD is a median, BH an altitude and CE an angle bisector.  They meet at G.

I then measured BA, CA and BC and calculated the two expressions.  As you can see they results are not equal.

Please check the wording of the question.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#4 2015-08-23 12:38:04

Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,956
Website

Re: A triangle has concurrent median, altitude and angle bisector

I like the OP's discrete optimization problems better.


'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.

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#5 2015-08-26 02:20:20

phanthanhtom
Member
Registered: 2012-06-22
Posts: 290

Re: A triangle has concurrent median, altitude and angle bisector

The thing is I'm better with discrete problems than geometry, and so I rarely gets the solution.

I have checked the source again. It seems to actually be: (BC + CA)(BC^2 + CA^2 - AB^2) = 2*BC*CA^2. (First term is BC not BA).

Sorry for the fault!

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#6 2020-08-24 06:57:03

MassaGeom
Member
Registered: 2020-08-24
Posts: 1

Re: A triangle has concurrent median, altitude and angle bisector

phanthanhtom wrote:

The thing is I'm better with discrete problems than geometry, and so I rarely gets the solution.

I have checked the source again. It seems to actually be: (BC + CA)(BC^2 + CA^2 - AB^2) = 2*BC*CA^2. (First term is BC not BA).

Sorry for the fault!


The simple proof of (BC+CA)(BC^2+CA^2-AB^2)=2*BC*CA^2:
1) AC/CB=AL/LB - angle bisector theorem
2) AL/LB*BD/DC*CH/AH - Cevas theorem for ABC and AD, BH, CL; BD=DC => AL/LB=AH/HC
3) CH=BC*cosC - because BHC=90
4) {1), 2), 3)} =>AC/CB=AL/LB=AH/HC=AH/(CB*cosC) => AC*cosC=AH => (AH+HC)*cosC=AH => (1+CH/AH)cosC=1.
cosC=(AC^2+BC^2-AB^2)/(2*AC*BC) and CH/AH=CB/AC (by {1), 2)}) => (1+CB/AC)*(AC^2+BC^2-AB^2)/(2*AC*BC)=1 =>
(BC+CA)(BC^2+CA^2-AB^2) = 2*BC*CA^2. Done!

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