ABC is a triangle such that AB = AC. Let D be the foot of the perpendicular from C to AB and E the foot of the perpendicular from B to AC.

Question

ABC is a triangle such that AB = AC. Let D be the foot of the perpendicular from C to AB and E the foot of the perpendicular from B to AC. Then

(a) $BC^3 < BD^3 + BE^3$

(b) $BC^3 = BD^3 + BE^3$

(c) $BC^3 > BD^3 + BE^3$

(d) None of the foregoing statements need always be true.

I drew the diagram and showed that the triangles BEC and BDC are similar and: $CE/BD=BE/CD=BC/BC=1$, or, $CE=BC$ and $BE=CD$

Then, from the same triangles, we know, $BD+CD>BC$, or, $BD+BE>BC$ (1) since $CD=BE$

We also know that $(BD-BE)^2>0$

or, $BD^2 -BE.BD+BE^2> BE.BD>0$ (2)

therefore from (1) and (2), I concluded that, $BE^3+BD^3>BC$

How do I proceed further?

Answer 1

Note that, triangles BDC and CEB are congruent as @SS_C4 mentioned. Then $CD = BE$, Now

$BC^2 = BD^2+CD^2$,

So $BC^3 = BD^{2}BC+CD^{2}BC = BD^{2}(BD+d_1)+CD^{2}(CD+d_2)\\ = BD^3+CD^3+BD^{2}d_1 + CD^{2}d_2 $

so $BC^3 > BD^3+CD^3$

Answer 2

Let $$AB=AC=b,BC=a$$ and $$\angle{ABC}=\angle{ACB}=\beta$$ so we get $$BD=\frac{a^2}{2b}$$ and $$BE=\frac{a}{b}\sqrt{b^2+\frac{a^2}{4}}$$ Now we show that $$BC^3-BD^3>BE^3$$ This is equivalent to $$a^3-\left(\frac{a^2}{2b}\right)^3>\left(\frac{a}{b}\sqrt{b^2-\frac{a^2}{4}}\right)^3>0$$ Squaring and factorizing gives $$(a^2+4ab+6b^2)(a-2b)^2>0$$ which is true.