Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to solve the following multiple choice problem:

$ABC$ 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

  1. $BC^3<BD^3+BE^3$
  2. $BC^3=BD^3+BE^3$
  3. $BC^3>BD^3+BE^3$
  4. none of the foregoing statements need always be true.

I have tried the following steps:

enter image description here

Since $BC$ is the hypotenuse for the right angled triangles $BCD$ and $BCE$, so $BC>BE$ and $BC>BD$, so that $BC^3>BE^3$ and $BC^3>BD^3$. Thus, $BC^3>\dfrac{BD^3+BE^3}{2}$. But the answer given in my textbook is option $3$ i.e. $BC^3>BD^3+BE^3$. But I can't get it. i need some help in this regard.

share|cite|improve this question
up vote 1 down vote accepted

By symmetry we have $BD=CE$. Now from the $BCE$ right triangle we have $$BC^2=CE^2+BE^2=BD^2+BE^2$$

Hence $BC^3=BC\times BD^2+BC\times BE^2>BD^3+BE^3$ (because $BC>BD$ and $BC>BE$).

share|cite|improve this answer
thanks a lot ! I got it now ! – Debashish Jun 30 '14 at 8:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.