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!

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!

]]>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'.)

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

]]>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.

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

]]>Please provide detailed proof.

]]>