Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question

1 Answer 1

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

Your Answer

 
discard

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.