Let $\Delta $ $ ABC$ be an arbitrary triangle, draw the altitudes $BD$ and $CF$ prove that $BC^2 = AB*FB + AC*DC$

Question

Let $\Delta$ $ ABC$ be an arbitrary triangle, draw the altitudes $BD$ and $CF$ prove that $BC^2 = AB*FB + AC*DC$

I have proven that $\Delta $ AFC $\sim $$ \Delta$ ADB using AA criterion.

so $\frac{AF}{AD} = \frac{AC}{AB}= \frac{FC}{DB}$ Also since $\Delta $ $FBC$ and $ \Delta$ $BCD$ are right triangles, we get $BC^2 = FC^2 + FB^2$ and $BC^2 = DB^2 + DC^2$

the other equality I got is $AB * CF = AC * BD$, it comes from the formula of the area of a triangle.

I tried something like this $$ AB * FB + AC * DC = AC(\frac{BD}{CF} FB + DC) $$ but after that I don't know how to continue, if I replace the fraction with $AB/AC$ I get the initial equation. I think I need to obtain $FC^2 + FB^2$ on the right hand side but I don't know how.

Any help is appreciated

Answer 1

Now, we need to prove that: $$BC^2=AB(AB-AF)+AC(AC-AD)$$ or $$BC^2=AB^2+AC^2-2AB\cdot AF,$$ which we can prove by the following way.

Let $F$ be placed between $A$ and $B$.

Thus, since $$CF^2=AC^2-AF^2=BC^2-BF^2,$$ we obtain $$BC^2-AC^2=BF^2-AF^2=(BF-AF)(BF+AF)=$$ $$=(AB-2AF)AB=AB^2-2AB\cdot AF,$$ which gives $$BC^2=AB^2+AC^2-2AB\cdot AF$$ and we are done!

Answer 2

It can be proved by cosine rule, In $\Delta ABC$, $$AC^2=AB^2+BC^2-2(BC)\cdot(AB)\cos\measuredangle B$$ $$=AB^2+BC^2-2(AB)\cdot (BF) \text{ $\because\cos\measuredangle B=\frac{BF}{BC}$}\tag{1}$$ Again in $\Delta ABC$ for side AB, $$AB^2=AC^2+BC^2-2(AC)(CD)\tag{2}$$ Adding $\text{eq.}(1)\text{ and }\text{eq.}(2),$ $$\require{cancel} \bcancel{AC^2}+\bcancel{AB^2}=\bcancel{AC^2}+BC^2+\bcancel{AB^2}+BC^2-2BF\cdot AB-2AC\cdot CD$$ $$\fbox{$BC^2=AB\cdot FB +AC\cdot DC$}$$