Finding values of dot products from a pyramid

Question: In the pyramid $ABCDE$, the base is a square with a side of length $5$ and $\vec{AD} \perp \vec{DE}$ . The vector $\vec{AE}$ creates equal angles with the vectors $\vec{AD}$ and $\vec{AB}$. Let us denote $\vec{AB}=u$, $\vec{AD}=v$ and $\vec{AE}=w$ (See the figure):

(a.) Showing all working, find the numerical value of $w⋅ v$ and the numerical value of $w⋅ u$

(b.) Show that $EDC$ is a right-angle triangle.

What I have done to start (a) we have that

$$\begin{cases} ||u||=5 \\ ||v||=5 \\ ||w||=? \\ \cos(\theta)=\frac{\vec{w}\cdot \vec{v}}{||w||||v||} \\ \cos(\theta)=\frac{\vec{w} \cdot \vec{u}}{||w||||u||} \\ \end{cases}$$

So since AE creates equal angles with AD and AB we can equate the angle of dot product:

$$\cos(\theta) = \cos(\theta) \Longrightarrow \frac{\vec{w}\cdot \vec{v}}{||w||||v||}=\frac{\vec{w} \cdot \vec{u}}{||w||||u||}$$

Then i don't know how to continue :(

I dont know at the moment a proof for b but in a take origin at $A$ from triangle rule we have $AE=AD+DE$ we want $AD.AE=AD (AD+DE)=|AD|^2+0=25$ as $AD$ is perpendicular to $DE$
you know that $\underline v\cdot\vec{DE}=0$ so the way to find $\underline w$ goes like this: $$\underline w=\underline v+\vec{DE}\\\underline v\cdot\underline w=\underline v\cdot\left(\underline v+\vec{DE}\right)\\\underline v\cdot\underline w=\underline v\cdot\underline v+\underline v\cdot\vec{DE}\\\underline v\cdot\underline w=\underline v\cdot\underline v\\\underline v\cdot\underline w=\|\underline v\|\|\underline v\|\cos0\\\underline v\cdot\underline w=\|v\|^2\\\underline v\cdot\underline w=25$$ for $\underline u\cdot\underline w\,$ you are doing the same and get the same result.
for (b.) you have the following:$$\vec{EC}=-\underline w+\underline v+\underline u\\\implies\vec{EC}\cdot\underline u=\left(-\underline w+\underline v+\underline u\right)\cdot\underline u\\=-\underline w\cdot\underline u+\underline v\cdot\underline u+\underline u\cdot\underline u\\=-\|\underline u\|\|\underline w\|+0+\|u\|^2\\=-25+25\\=0$$ and you know that $\vec{EC}\cdot\underline u=\|\vec{EC}\|\|\underline u\|\cos\theta,\|\vec{EC}\|\ne0\ne\|\underline u\|\,$ therefore $\theta=90$