Question

Prove that if ABC is triangle and Dis the mid point of BC , then AB^2+AC^2,=2*AD^2+DC^2)

Answer 1

Answer on Question #67211 - Math - Analytic Geometry

Question

Prove that if ABC is triangle and Dis the mid point of BC, then $\mathrm{AB}^{\wedge}2 + \mathrm{AC}^{\wedge}2, = 2^{*}(\mathrm{AD}^{\wedge}2 + \mathrm{DC}^{\wedge}2)$

Solution

Draw in triangle $ABC$ segment $AD$ . Since $D$ is the midpoint of $BC$ , so $AD$ is the median from $A$ to $BC$ . Consider triangle $ABD$ . Use the Law of cosines (or cosine rule). We have

$$A D ^ {2} = A B ^ {2} + B D ^ {2} - 2 A B \cdot B D \cdot \cos \alpha$$

Consider triangle $ACD$ . Use the Law of cosines. We have

$$A D ^ {2} = A C ^ {2} + C D ^ {2} - 2 A C \cdot C D \cdot \cos \beta$$

Add (1) and (2):

$$2 A D ^ {2} = A B ^ {2} + B D ^ {2} + A C ^ {2} + C D ^ {2} - 2 A B \cdot B D \cdot \cos \alpha - 2 A C \cdot C D \cdot \cos \beta$$

Take into account that $BD = DC$ . Then we get

$$2 A D ^ {2} = A B ^ {2} + A C ^ {2} + 2 D C ^ {2} - 2 D C (A B \cdot \cos \alpha + A C \cdot \cos \beta)$$

Now we draw in triangle $ABC$ the altitude $AE$ . We get two right triangles $ABE$ and $ACE$

In triangle $ABE$ we have

$$B E = A B \cdot \cos \alpha$$

and in triangle $ACE$ we have

$$C E = A C \cdot \cos \beta$$

Substituting (4) and (5) into (3) we get

$$2 A D ^ {2} = A B ^ {2} + A C ^ {2} + 2 D C ^ {2} - 2 D C (B E + C E)$$

however $BE + CE = BC = 2DC$ . Then for (6) we get

$$2 A D ^ {2} = A B ^ {2} + A C ^ {2} - 2 D C ^ {2}$$

And finally

$$A B ^ {2} + A C ^ {2} = 2 \left(A D ^ {2} + D C ^ {2}\right)$$

Answer: if ABC is triangle and D is the mid point of BC, then

$$A B ^ {2} + A C ^ {2} = 2 \left(A D ^ {2} + D C ^ {2}\right)$$

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