ABC is an equilateral triangle it's center is m , straight line (L) was drawn passing through the point m and in the plane of the triangle. Prove that "The sum of heights lengths squared drawn from the heads of the triangle on the line (L) equal a constant whatever the line (L) was."

- TrojanPoem

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- ganeshie8

the question is not so clear to me, could you draw it out showing what you mean by "heights lengths" and "heads of triangle" ?

- anonymous

@dan815 Why did u leave me
;(

- TrojanPoem

head of triangle means A, B, C . Heights lengths : Like drawing perpendicular lines to the line L

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- TrojanPoem

@dan815 , Go help him dan

- dan815

|dw:1435577825081:dw|

- ganeshie8

got it, thnks

- dan815

|dw:1435578006968:dw|

- dan815

|dw:1435578050263:dw|

- ganeshie8

is the constant same as side length ?

- ganeshie8

|dw:1435578155241:dw|

- dan815

oo nice

- dan815

limiting case

- perl

*

- dan815

|dw:1435578420056:dw|

- TrojanPoem

Wait I did a mistake , the sum of height lengths squared

- dan815

|dw:1435578673892:dw|

- dan815

kk

- dan815

is there an elegant solution, not liek a calculus solution

- dan815

cuz you can alwys write a calculus solution just rote

- dan815

if its true

- TrojanPoem

Lol , the solution isn't calculus related dan815, It all depends on geometry and there is no rare formulas

- dan815

y=mx+b and write perpendicular equations from 3 equal points

- ganeshie8

then i guess areas/similar triangles works pretty well
|dw:1435578883770:dw|

- TrojanPoem

@dan815, I though you'd get it easily. It's just lengths calculations

- TrojanPoem

@ganeshie8 , How ?

- ganeshie8

idk how, im still thinking.. let me grab pen paper

- dan815

fomr the similar triangles we can see the hypotenuses ratios are equal i can see thats where the square relationship is popping out

- dan815

|dw:1435579184896:dw|

- ikram002p

dan so much typos u made :O

- dan815

my writing gives peoples unknowns to solve

- dan815

people*

- ikram002p

i didnt got this
"of heights lengths squared drawn from the heads of the triangle on the line (L) equal a constant whatever the line (L) was."
@ganeshie8 could u give an example ?

- dan815

|dw:1435579873776:dw|

- dan815

it says a^2+b^2+c^2 = k

- ikram002p

is this a case ?
|dw:1435580036617:dw|

- dan815

|dw:1435579995219:dw|

- dan815

|dw:1435580085369:dw|

- ikram002p

ok so L is parallel to one side and pass the center ?( the only case allowed ?)

- ikram002p

hmm i seems dont understand :)

- ikram002p

hello xD

- dan815

its turning into an algebra excercise on my end, am i on the wrong path then

- dan815

a^2 , b^2 and c^2 im relating them all to their hypotenuses which i know are constant and now we know the rations are same and we have different relationships for the hypotenuse as well

- ikram002p

i just wanna see what about this case ?
|dw:1435581045995:dw|

- dan815

|dw:1435581053518:dw|

- dan815

|dw:1435581167187:dw|

- ikram002p

ugh "sigh"

- dan815

what about that diagram?

- dan815

it's all legal, just draw any line through m and show the sum of the squares of the perpendiculars from the corners of the triangle to that line through m, which is the center of the triangle are equal

- TrojanPoem

@ikram002p , The line (L) is not parallel to the side.

- TrojanPoem

|dw:1435581369777:dw|

- ikram002p

ok we know all right triangles are congruent for some cool reason :D

- ikram002p

|dw:1435582302688:dw|

- ganeshie8

I have managed to show \(a+b=c\)

- ganeshie8

|dw:1435582837097:dw|

- ganeshie8

feel free to use that if it helps,
il provide the proof in the end..

- ikram002p

did u use similarities properties ?

- ikram002p

if not i'll continue with mine lol if yes i'll take ur result

- ganeshie8

continue with yours..

- ikram002p

k :)

- ikram002p

\(c/b=(BH)/(HC)=\beta \\
c/a=(BG)/(GA)=\alpha \\ c=\alpha a,c=\beta b---2\text{ equations i'll use also ganeshie8 result }\\ a+b=c\\ \)
\((\alpha -2)^2a^2+(\beta-2)^2b^2+\frac{2(\alpha-2)(\beta-2)}{\alpha \beta }c^2=0\)
|dw:1435585962332:dw|

- TrojanPoem

The constant is 1/2 The length of the triangle side squared.

- TrojanPoem

I don't think that the three triangles are similar. Even if they are. To use similarities It must be proven.

- ikram002p

I have shown the similarty case, not all three are similar

- TrojanPoem

So the solution was build on corrupted base ?

- dan815

|dw:1435666204039:dw|

- ganeshie8

|dw:1435669635846:dw|

- ikram002p

this is amazing !!!

- ganeshie8

Here is a dumb proof using coordinate geometry which of course i don't like :
Let
\(A = (-t,0)\)
\(B=(t,0)\)
\(C=(0, \sqrt{3}t)\)
|dw:1435675544583:dw|

- ganeshie8

Centroid is clearly \((0, \frac{t}{\sqrt{3}})\), call this point \(G\).
Then any line passing through this point is given by \[y=mx+\frac{t}{\sqrt{3}}\tag{1}\]
|dw:1435676089356:dw|

- ganeshie8

Sum of squares of distances from the line\((1)\) to the vertices :
\[\begin{align}
&\dfrac{(tm+\frac{t}{\sqrt{3}})^2+(-tm+\frac{t}{\sqrt{3}})^2+(-\sqrt{3}t+\frac{t}{\sqrt{3}})^2}{m^2+1}\\~\\
&=\dfrac{2(t^2m^2+\frac{t^2}{3}) +\frac{4t^2}{3}}{m^2+1}\\~\\
&=\dfrac{2(t^2m^2+t^2)}{m^2+1}\\~\\
&=2t^2
\end{align}\]

- TrojanPoem

Done ?

- ganeshie8

you don't get it ?

- TrojanPoem

so the constant is .... from the triangle side ?

- ganeshie8

|dw:1435677076734:dw|

- TrojanPoem

yeah, got it.

- TrojanPoem

Thanks ganeshie! xD

- ganeshie8

im curious to see your proof

- TrojanPoem

It's using angles

- ganeshie8

because i don't like my proof
i don't like any proof that uses coordinate geometry

- ganeshie8

please do share

- TrojanPoem

Ok.I will write it.

- TrojanPoem

|dw:1435677412136:dw|

- TrojanPoem

<(amk) = <(amb) - <(bmk) = 120 - (180 - w) = w - 60
<(cmn) = <(cmb) - w = 120 -w
Assume that ma = mb = mc = y
in triangle amk:
ak = ma * sin(w- 60) = y * sin(w-60)
(ak)^2 = y^2 [sin(w-60)]^2
in triangle bmh:
bh = bm * sinw
(bh)^2 = bm^2 [sinw]^2

- TrojanPoem

bh^2 = y^2 [sinw]^2
in triangle cmn
cn = mc * sin(120 -w) = y sin (w +60)
cn^2 = y^2 sin^2[w +60]
Sum cn^2 + bh^2 + ak^2

- TrojanPoem

y^2sin(w-60)^2 + y^2 sin^2(w) + y^2sin(w+60)^2
= y^2(sin(w-60)^2 + sin^2(w) + sin(w+60)^2)
Now let's calculate sin^2(w-60) = [ sinw cos 60 - sin60 cosw]^2
= [0.5 sinw - 0.5sqrt(3)cosw]^2

- TrojanPoem

= 0.25 sin^2(W) - 0.5sqrt(2)sinwcosw + 3/4 cos^2(w)

- TrojanPoem

sin(w + 60)^2 = 0.25sin^2(w) + 0.5sqrt(3) sinwcosw + 3/4 cos^2(w)

- TrojanPoem

cn^2 + bh^2 + ak^2 ( put the values here)
(3/2 sin^2 w + 3/2 cos^2w) * y^2
assume the length of the triangle side is x
x = sqrt(3) y
so y = x/sqrt(3)

- TrojanPoem

put the values 3/2 * (sin^w+cos^2w) y^2
3/2 * (x/sqrt(3))^2 = 0.5 x^2

- ganeshie8

gotcha! happy to see an alternative but i don't really like this either as it is more of algebra/trig and doesn't provide much insight on geometry
last night i was trying to work it w/o trig using only similar triangles and other euclid stuff

- TrojanPoem

Yeah, It's all about angles here. Take harder one :) ?

- ganeshie8

Sure :)
Prove this : \(a+b=c\)
where \(a,b,c\) are the perpendicular distances from vertices to a line passing through centroid of an equilateral triangle as shown :
|dw:1435678372496:dw|