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
We consider the picture:
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... $$ \begin{aligned} &c^2 = \color{red}{h^2}+x^2\ ,\\ &b^2 = \color{red}{h^2}+y^2\ ,\\ &s^2 %= \color{red}{h^2}+(y-m)^2\\ = \color{red}{h^2}+(x-n)^2\ ,\\[3mm] &AB^2\cdot m + AC^2\cdot n \\ &\qquad=c^2m +b^2n \\ &\qquad=(\color{red}{h^2}+x^2)m +(\color{red}{h^2}+y^2)n\ , \\[3mm] &AD^2\cdot (m+n)\\ &\qquad=s^2(m+n)\\ &\qquad=(\color{red}{h^2}+DE^2)(m+n)\\ &\qquad=(\color{red}{h^2}+(x-n)(m-y))(m+n)\\[3mm] &BD^2\cdot m + DC^2n+AD^2\cdot (m+n)\\ &\qquad=n^2m + m^2n + (\color{red}{h^2}+(x-n)(m-y))(m+n)\\ &\qquad=\color{red}{h^2}(m+n) +mn(m+n)+(xm+yn-mn-xy)(m+n)\\ &\qquad=\color{red}{h^2}(m+n)+(xm+yn-xy)(m+n)\\ &\qquad=\color{red}{h^2}(m+n)+(xm+yn-xy)(x+y)\\ &\qquad=\color{red}{h^2}(m+n)+(x^2m+y^2n) \\ &\qquad\qquad+\underbrace{(xym+xyn-x^y-xy^2)}_{=xy(m+n-x-y)=xy(a-a)=0}\\ &\qquad=\color{red}{h^2}(m+n)+(x^2m+y^2n) \\ &\qquad=(\color{red}{h^2}+x^2)m+(\color{red}{h^2}+y^2)n\\ &\qquad=c^2m+b^2n\\ &\qquad=AB^2\cdot m+AC^2\cdot n \ . \end{aligned} $$
$\square$