Vectors Question about proving if sides of square are perpendicular.

Question

The squares $ACGF$ and $B C D E$ are constructed on the sides of a triangle $A B C,$ Let $a=\overrightarrow{C A}$, $b=\overrightarrow{C B}, u=\overrightarrow{G C}$ and $v=\overrightarrow{D C} .$ Let $T$ be the midpoint of $A B$ (see the diagram).

(a) Write $\overrightarrow{G D}$ in terms of $u$ and $v$.

(b) Write $\overrightarrow{C T}$ in terms of $a$ and $b$.

(c) Calculate $u \cdot a$. Explain your reasoning.

(d) Show that $\overrightarrow{G D} \cdot \overrightarrow{C T}=\frac{1}{2}(u \cdot b-v \cdot a)$ Provide your reasoning for each step.
Hint: Use your answers from $(\mathrm{a})-(\mathrm{c})$ to expand $\overrightarrow{G D} \cdot \overrightarrow{C T}$ It is known that the angle between the vectors $u$ and $b$ is equal to the angle between the vectors $v$ and $a$ (you do not need to prove this).

(e) Use the fact above and the result from part (d) to prove that $\overrightarrow{G D} \perp \overrightarrow{C T}$. Provide your reasoning for each step.

Original Question

Answer 1

Let $H $ be such point that $ACBH $ is paralelogram. Observe rotation around A for $90^{\circ}$. It takes $C$ to $F$ and $H$ to new point $H'$. Then $BH'DC$ is also paralelogram since $AH'=CD$ and $AH'||CD$. Thus $\triangle GCD \cong \triangle FAH'$ so $GD||FH'$ (translation for vector $\vec{AC}$ takes $FH'$ to $GD$). Q.E.D.

Answer 2

$\vec{GD } = u-v$ and $\vec{CT } = {1\over 2}(a+b)$

Since $a$ and $u$ are perpendicular we have $a\cdot u=0$ and the same $v\cdot b=0$.

Then

$$\vec{GD } \cdot \vec{CT } = {1\over 2}(u-v)(a+b) = {1\over 2}(ua+ub-va-vb) = {1\over 2}(ub-va)$$

Finally $$u\cdot b = ||u||\cdot ||b|| \cdot \cos (90^{\circ}+\gamma)$$ and

$$v\cdot a = ||v||\cdot ||a|| \cdot \cos (90^{\circ}+\gamma)$$

But $||v||= ||b||$ and $||u||= ||a||$ so $u\cdot b = v\cdot a$ and thus $GD \bot CT$ Q.E.D.