How to prove that $\sqrt[3]{BC^2}=\sqrt[3]{BD^2}+\sqrt[3]{CE^2}$.

Question

So we have ABC is a right triangle at A, AH be the altitude, HD, HE are respectively the height of the triangle AHB and AHC. Prove that $\sqrt[3]{BC^2}=\sqrt[3]{BD^2}+\sqrt[3]{CE^2}$.

I try using Pythagorean theorem but I only end up at $BC^2=3(AH^2)+BD^2+CE^2$ and I don't know where to go next or maybe I choose wrong path I also try to use what we're proving and simplify it too but no luck. It looks kinda easy but I've just started studying Geometry ( it's still difficult for me )so I hope everyone here could help me, Thanks

score 4 · Answer 1 · answered Sep 06 '19 at 12:56

4

Notice $\triangle ABC \sim \triangle HBA \sim DBH$, we have $$\frac{AB}{BC} = \frac{HB}{BA} = \frac{DB}{BH} \implies \frac{DB}{BC} = \left(\frac{AB}{BC}\right)^3 $$ Similarly, $\triangle ABC \sim \triangle HAC \sim \triangle EHC$ leads to $$\frac{AC}{BC} = \frac{HC}{AC} = \frac{EC}{HC}\implies \frac{EC}{BC} = \left(\frac{AC}{BC}\right)^3$$ This implies $$\sqrt[3]{\frac{DB^2}{BC^2}} + \sqrt[3]{\frac{EC^2}{BC^2}} = \frac{AB^2 + AC^2}{BC^2}$$ Since $\angle CAB = 90^\circ$, by Pythagorean theorem, RHS is $1$ and we are done.

answered Sep 06 '19 at 12:56

achille hui

122,701

Thank you so much for your help now I could sleep in peace haha – user635988 Sep 06 '19 at 13:01
Hmm...sorry for asking you to explain me, but I try to see the solution and find some I couldn't understand at all. How do you get $$ \frac{DB}{BC} = \left(\frac{AB}{BC}\right)^3 $$ – user635988 Sep 06 '19 at 13:19
@user635988 $\frac{DB}{BC} = \frac{DB}{BH}\frac{BH}{AB}\frac{AB}{BC}$. – achille hui Sep 06 '19 at 13:22

How to prove that $\sqrt[3]{BC^2}=\sqrt[3]{BD^2}+\sqrt[3]{CE^2}$.

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