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Question - In triangle ABC, point D divides BC in n/m(=BD/DC) ratio. AE is the altitude from A to BC. Prove that (AB^2)*m + (AC^2)n = m(BD^2) + n*(CD^2) + (m+n)*(AD^2)

I was trying to prove Stewart's theorem by simply expanding LHS(AB^2*m + AC^2*n) and using Pythagoras theorem in some places. I got the proof till the end, but I only have left to prove that (ED^2 + BE^2)/BD^2 = n

skallu
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  • Please note that your question is not connected to showing, or disproving the relation $(ED^2+BE^2)/BD^2=n$, but rather with the proof of the Stewart's Theorem, where the idea of proof is given. This relation is "false" in the sense that it is not even well defined. To be well defined, it has to be homogeneous in $n,m$ since both variables $n,m$ do not have a meaning per se, but we can give sense to a relation between then (only) when they appear in the same monomial degree. If you want a check of own computations, insert them. Best using mathjax. (Answers use it, yes this is work.) – dan_fulea Apr 15 '20 at 11:03

1 Answers1

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We consider the picture:

Stewart's Theorem

and want to show the following relation (Stewart) by using the point $E$ as an intermediate point. $$ AB^2\cdot m + AC^2\cdot n = BD^2\cdot m + CD^2\cdot n + AD^2\cdot (m+n)\ . $$ Here, $n=BD$, $m=DC$, and we will use the notations $x=BE$, $y=EC$, $h=AE$, $s=AD$, and the usual notations $a=BC=m+n=x+y$, $b=CA$, $c=AB$. Then we have: $$ \begin{aligned} &c^2 = \color{red}{h^2}+x^2\ ,\\ &b^2 = \color{red}{h^2}+y^2\ ,\\ &s^2 %= \color{red}{h^2}+(y-m)^2\\ = \color{red}{h^2}+(x-n)^2\ ,\\ &AB^2\cdot m + AC^2\cdot n - AD^2\cdot (m+n)\\ &\qquad=c^2m +b^2n -s^2(m+n)\\ &\qquad=(\color{red}{h^2}+x^2)m +(\color{red}{h^2}+y^2)n -(\color{red}{h^2}+(x-n)^2)(m+n)\\ &\qquad=x^2m+y^2n-(x-n)^2(m+n)\\ &\qquad=x^2(a-n)+(a-x)^2n-(x-n)^2a\\ &\qquad=(x^2a-x^2n)+(a^2n-2axn+x^2n)-(x^2a-2axn+n^2a)\\ &\qquad=a^2n-n^2a=an(a-n)=amn=(m+n)mn=m^2n+n^2m\\ &\qquad=BD^2\cdot m + DC^2n\ . \end{aligned} $$

$\square$

Later EDIT: The above is for me the simpler way to display and remember the calculus, but ok, it is easy to rewrite it as wanted in the comments. If this helps... $$ \begin{aligned} &c^2 = \color{red}{h^2}+x^2\ ,\\ &b^2 = \color{red}{h^2}+y^2\ ,\\ &s^2 %= \color{red}{h^2}+(y-m)^2\\ = \color{red}{h^2}+(x-n)^2\ ,\\[3mm] &AB^2\cdot m + AC^2\cdot n \\ &\qquad=c^2m +b^2n \\ &\qquad=(\color{red}{h^2}+x^2)m +(\color{red}{h^2}+y^2)n\ , \\[3mm] &AD^2\cdot (m+n)\\ &\qquad=s^2(m+n)\\ &\qquad=(\color{red}{h^2}+DE^2)(m+n)\\ &\qquad=(\color{red}{h^2}+(x-n)(m-y))(m+n)\\[3mm] &BD^2\cdot m + DC^2n+AD^2\cdot (m+n)\\ &\qquad=n^2m + m^2n + (\color{red}{h^2}+(x-n)(m-y))(m+n)\\ &\qquad=\color{red}{h^2}(m+n) +mn(m+n)+(xm+yn-mn-xy)(m+n)\\ &\qquad=\color{red}{h^2}(m+n)+(xm+yn-xy)(m+n)\\ &\qquad=\color{red}{h^2}(m+n)+(xm+yn-xy)(x+y)\\ &\qquad=\color{red}{h^2}(m+n)+(x^2m+y^2n) \\ &\qquad\qquad+\underbrace{(xym+xyn-x^y-xy^2)}_{=xy(m+n-x-y)=xy(a-a)=0}\\ &\qquad=\color{red}{h^2}(m+n)+(x^2m+y^2n) \\ &\qquad=(\color{red}{h^2}+x^2)m+(\color{red}{h^2}+y^2)n\\ &\qquad=c^2m+b^2n\\ &\qquad=AB^2\cdot m+AC^2\cdot n \ . \end{aligned} $$

$\square$

dan_fulea
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  • Can you please solve it without moving AD^2(m+n) to the LHS – skallu Apr 12 '20 at 12:36
  • "Moving $AD^2(m+n)$ to the other side", and also setting $\color{red}h$ in this red color, were two steps done for the reason that after building $AB^2\cdot m+AC^2\cdot n-AD^2\cdot (m+n)$ the $\color{red}h$-contribution cancel each other, and we are left with an algebraic expression in $x,y;m,n$, which all represent "data related to pieces on $BC$", thus to data in "one dimension". This was the leading idea behind the computations, and this is also the reason why this kind of computation works. (And this is also a possibility to search for the Stewart relation.) For me it's simpler so. But... – dan_fulea Apr 13 '20 at 14:48
  • i've added also an other way to lead the computation, also without breaking the "symmetry" at each step. The terms in the LHS will stay always together. – dan_fulea Apr 13 '20 at 14:50
  • I tried to expand AB^2m+AC^2n and found that I must prove (ED^2 + BE^2)/BD^2 = n to prove Stewarts theorem. Please check if this is true. – skallu Apr 15 '20 at 08:53
  • This is the last touch of this subject. Note that $n$ is not defined (in the setting in the OP), but rather defined up to a factor. Only the proportion $n:m$ is defined. We can take $n=BD$, $m=DC$, as it is the case in the classical formulation of Stewart's Theorem. Now a formula of the shape $(ED^2+BE^2)/BD^2=n$ is false for the reason of "bad dimensionality". (If segments are measured in [cm], then on the LHS we have an adimensional quantity, on the RHS $n=BD$ measured in [cm].) All relations i write have degree $3$. Check your computations for homogenity. (We do not have your computations) – dan_fulea Apr 15 '20 at 10:57