Let $c(t), \bar{c}(t)$ be two intersecting smooth paths in a smooth surface $S$, lying entirely in a chart domain $D$, and define smooth paths $w(t) = φ ◦ c(t), \bar{w}(t) = φ ◦ \bar{c}(t)$ in $U$, where $φ: D → U ⊂ \mathbb{R} ^{2}$ is the chart map. Prove that the angle of intersection between $w(t)$ and $\bar{w}(t)$ is the same as that between $c(t)$ and $\bar{c}(t)$, for all such paths in $D$, if and only if the chart is isothermal.
I have proved one way of the iff and I need to prove given $\theta= \phi$ are the angles of intersection then the chart is isothermal.
I have taken $w(t) = (u(t), v(t))$ and $\bar{w} = (\bar{u}(t), \bar{v}(t))$ so that we have $c'=p_{u}u'+p_{v}v'$ and $\bar{c}'=p_{u}\bar{u}'+p_{v}\bar{v}'$ for $p= \varphi^{-1}$.
Taking $u′ = 1, $ $v ′ =0$ and $\bar{u}′ = 0,$ $ \bar{v}′ = 1$ I have $$\frac{\langle c',\bar{c}'\rangle}{|c'||\bar{c}'|}=\frac{F}{\sqrt{EG}}=cos(\theta)=cos(\phi)$$.
I believe then this should be equal to $0$ to show that $F=0$ and the chart is orthogonal. I can then prove in a similar way that $E=G$ however I am unsure why this would be equal to $0$.
HINT: This is just a linear algebra exercise. Show that a linear map $T = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ preserves angles if and only if $a=d$ and $b=c=0$. In particular, if the unit square maps to a rectangle, think about angles with the diagonal.