Prove the Apollonius' theorem.

Question

Let in a $\Delta ABC$, D is the midpoint of $BC$.Prove that:

$AB^2+AC^2=2CD^2+2AD^2$

MY ATTEMPT :

Given that $BD=DC$ and we construct $E \ such \ that\ AE=EC\implies AC=\frac{EC}{2} \ and \ DE||AB \implies DE=\frac{AB}{2}$

For the $\Delta DEC$ we have $DC^2=DE^2+EC^2 \implies 4DC^2=AC^2+AB^2 $

we have $AB^2+AC^2=2CD^2+2CD^2 \tag{1}$

In $\Delta ADE$ we have, $AD^2=AE^2+DE^2 \implies AE^2+DC^2-EC^2 \implies AD^2=DC^2$

Hence $2CD^2=2AD^2$

Thus, we have

$AB^2+AC^2=2CD^2+2AD^2$

I am not sure of this proof. Though this proof is well explained in wikipedia.I tried to check if this can be solved using elementary geometry.

Why is $\angle DEC=90^o$? It may or may not be $90^o$. BTW, an alternate proof is given below. — Apurv, Apr 30 '15 at 16:15
Yes! I am not convinced with my proof. Your's is neat. Mine is just for a single case. — gaufler, Apr 30 '15 at 16:20

Apurv · Accepted Answer · 2016-02-17T13:22:23.833

Use the cosine rule twice:

$$\cos B=\dfrac {c^2+\frac {a^2}{4}-m^2}{2\times c\times \frac {a}{2}}\implies ac\cos B=c^2+\dfrac {a^2}{4}-m^2$$

$$\cos C=\dfrac {b^2+\frac {a^2}{4}-m^2}{2\times b\times \frac {a}{2}}\implies ab\cos C=b^2+\dfrac {a^2}{4}-m^2$$

Adding the above two and using $a=b\cos C+c\cos B$, we get $$a^2=b^2+c^2+\dfrac {a^2}{2}-2m^2$$ or $$AB^2 +AC^2 =2CD^2 +2AD^2$$(on rearranging the terms)

Prove the Apollonius' theorem.

1 Answers1