How to prove a length in a triangle when all three sides are known

Question

Question:

In $\Delta ABC$, the lengths of all three sides are known: $AB = c$, $BC = a$ and $AC = b$. Suppose a point $D$ is the foot of the height drawn from the vertex $C$ to the line $AB$.

Prove that $$AD = \dfrac{c^2 - a^2 - b^2}{2c}$$

I am thinking of solving the problem through somehow finding a relationship between the triangle as a whole and the two smaller triangles contained within. However, I am not sure how to being solving the question.

Answer 1

It isn't true. Take $a=b=c$ and receive $AD=-\frac a2$

The right formula is

$AD=b\cdot \cos A=b\cdot\frac{b^2+c^2-a^2}{2bc}=\frac{b^2+c^2-a^2}{2c}$

Answer 2

The statement you wrote is not correct. However, you can still derive the length by using the cosine law: $$\cos A = \frac{b^2+c^2-a^2}{2bc}$$ and $$AD = AC\cdot \cos A = \frac{b^2+c^2-a^2}{2c}$$ which I believe is the result you are looking for.

Answer 3

What you are looking at is the projection of $AC=b$ on side $BC$ having a projected length $ AD$ (red line).