1. ## Apollonius Theorem

Hi, i am having trouble with this proof. I have found mulitple webpages on the theorem, but not one of them i coudl find offers me a 'how to'

It is to do with Appolonis' Theorem. The question is:

"In triangle ABC, D is the midpoint of BC, then prove that -
AB² + AC² = 2AD² + 2CD²

Hint: Use Pythagoras' Theorem."

although from looking at the web, i am thinking my teacher has copied it down incorrectly since they all seem to say AB² + AC² = 2AD² + (BC²)/2

since AD is a median, and it is not known that there are any right angles in this triangle, i am not sure how to get Pythagoras in there at all

thanks for any help

jacs

2. You can use Stewart's theorem.

3. Hello, jacs!

In $\displaystyle \Delta ABC,\:D$ is the midpoint of $\displaystyle BC$.
Prove that: .$\displaystyle AB^2 + AC^2 \:= \:2\!\cdot\!AD^2 + 2\!\cdot\!CD^2$

We have: .$\displaystyle \Delta ABC$ with median $\displaystyle AD$.
Let $\displaystyle BD = DC = x$.

Drop perpendicular $\displaystyle AE$ from vertex $\displaystyle A$ to side $\displaystyle BC$. .Let $\displaystyle h = AE$.
.Let $\displaystyle a = ED$, then: $\displaystyle BE = x - a$
Code:
              A
*
*:* *
* : *   *
*  :  *     *
*   :   *       *
*    :h   *         *
*     :     *           *
*      :      *             *
B * - - - + - - - * - - - - - - - * C
x-a  E   a   D       x
In right triangle $\displaystyle AEB\!:\;\;AB^2\;=\;h^2 + (x-a)^2$ . [1]

In right triangle $\displaystyle AEC\!:\;\;AC^2 \;=\;h^2 + (x+a)^2$ . [2]

Add [1] and [2]: .$\displaystyle AB^2 + AC^2 \;=\;h^2 + (x-a)^2 + h^2 + (x+a)^2$

And we have: .$\displaystyle AB^2 + AC^2 \;=\;2(h^2+a^2) + 2x^2$ . [3]

. . In right triangle $\displaystyle AED\!:\;\;h^2 + a^2 \:=\:AD^2$
. . . . We also know that:.$\displaystyle x = CD$

Therefore, [3] becomes: .$\displaystyle AB^2 + AC^2 \;=\;2\!\cdot\!AD^2 + 2\!\cdot\!CD^2$

4. Thanks so much for that. I was of course going about it all wrong and now I have seen that, it is very clear.

Thanks for a concise explanation too, made it very easy to follow

5. Originally Posted by jacs
Hi, i am having trouble with this proof. I have found mulitple webpages on the theorem, but not one of them i coudl find offers me a 'how to'

It is to do with Appolonis' Theorem. The question is:

"In triangle ABC, D is the midpoint of BC, then prove that -
AB² + AC² = 2AD² + 2CD²

Hint: Use Pythagoras' Theorem."

although from looking at the web, i am thinking my teacher has copied it down incorrectly since they all seem to say AB² + AC² = 2AD² + (BC²)/2

since AD is a median, and it is not known that there are any right angles in this triangle, i am not sure how to get Pythagoras in there at all

thanks for any help

jacs
As D is the midpoint of BC, 2CD^2 = 2(BC/2)^2=(BC^2)/2.

RonL

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# 10th geometry problems on apollonious Theorem

Click on a term to search for related topics.