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This proof is found on page 172 of Sheldon Axler's Linear Algebra Done Right, Third Edition:

Considering the orthogonal decomposition:

$$u = \frac{\langle u, v\rangle}{\lVert v \rVert^2}v + w\tag 1$$

and since $v$ and $w$ are orthogonal, by PT:

$$\lVert u \rVert^2=\Bigg\lVert \frac{\langle u, v\rangle}{\lVert v \rVert^2}v \Bigg \rVert^2+\lVert w\rVert^2$$

The next step is the object of my question:

$$\lVert u \rVert^2=\frac{\langle u, v\rangle^2}{\lVert v \rVert^2}+\lVert w\rVert^2.$$

Equation 1 comes from a right triangle, where a leg is the scalar multiplication of a vector $v$ by $c$; the hypotenuse is $u$, and the other leg is $w = u - cv$:

$$u = cv + (u - cv)$$

To fulfill the perpendicular relationship of the legs,

$$0 = \langle u - cv, v \rangle=\langle u,v \rangle - c \lVert v\rVert^2$$

Hence, $c = \frac{\langle u,v \rangle}{\lVert v \rVert^2}$, and

$$u = \frac{\langle u,v \rangle}{\lVert v \rVert^2} v+\left(u - \frac{\langle u,v \rangle}{\lVert v \rVert^2}v\right)$$

where $w = u - \frac{\langle u,v \rangle}{\lVert v \rVert^2}v.$

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    What exactly is your question -- is it how to justify that step? – 2017-01-02
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    @littleO Yes. Embarrassingly simple. – 2017-01-02
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    No, that's not a bad question. – 2017-01-02

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$$\Bigg\lVert \frac{\langle u, v\rangle}{\lVert v \rVert^2}v \Bigg \rVert^2=\left(\frac{\langle u, v\rangle}{\lVert v \rVert^2}\right)^2\lVert v \rVert^2=\frac{\langle u, v\rangle^2}{\lVert v \rVert^2}$$