$\angle AOB = 30^o $, C is an arbitrary point inside the angle and OC =$1$. If we choose two points D and E on OA and OB to form $\Delta$CDE, what is its minimum perimeter? With very little information, I tried to draw perpendicular lines CD $\perp$ OA, and CE $\perp$ to OB, but it does not seem to work. Since C is an arbirtary point, need to find some symmetry somewhere?
Hint: Reflect $C$ across $OA$ and $OB$ as shown.
The perimeter $CD+DE+EC = C'D + DE + EC'_1$. When does the latter reach its minimum? It should then follow trivially that the minimal perimeter is $1$.
Edited: Note that I personally prefer Player3236's answer.
Consider $OC$ your baseline. Then $\angle DOC = 15^\circ$ and $\angle COE = 15^\circ$, symmetrically above and below $OC$. Due to symmetries, you can assume $DE \perp OC$.
In a coordinate system where $O = (0,0)$, $C = (1,0)$, $D = (x(t), y(t))$, and $E = (x(t), - y(t))$, the perimeter $p$ is $$p(t) = 2 y(t) + 2 \sqrt{ y(t)^2 + (x(t) - 1)^2 }$$ This reaches a minimum where its derivative reaches zero, $$\frac{d p(t)}{d t} = 2 \frac{d y(t)}{d t} + \frac{2 y(t)\left(\displaystyle \frac{d y(t)}{d t}\right) + 2 \left(x(t) - 1\right)\left(\displaystyle \frac{d x(t)}{d t}\right)}{\sqrt{y(t)^2 + (x(t)-1)^2}} = 0$$ You can use either $$\left\lbrace \begin{aligned} x(t) &= t \cos 15^\circ = t \displaystyle \frac{\sqrt{3} + 1}{2 \sqrt{2}} \\ y(t) &= t \sin 15^\circ = t \displaystyle \frac{\sqrt{3} - 1}{2 \sqrt{2}} \\ \end{aligned} \right. \Rightarrow \left\lbrace \begin{aligned} \displaystyle \frac{d x(t)}{d t} &= \cos 15^\circ = \displaystyle \frac{\sqrt{3} + 1}{2 \sqrt{2}} \\ \displaystyle \frac{d y(t)}{d t} &= \sin 15^\circ = \displaystyle \frac{\sqrt{3} - 1}{2 \sqrt{2}} \\ \end{aligned} \right. $$ or $$\left\lbrace \begin{aligned} x(t) &= t \\ y(t) &= t \tan 15^\circ = t \left( 2 - \sqrt{3} \right) \\ \end{aligned} \right. \Rightarrow \left\lbrace \begin{aligned} \displaystyle \frac{d x(t)}{d t} &= 1 \\ \displaystyle \frac{d y(t)}{d t} &= \tan 15^\circ = 2 - \sqrt{3} \\ \end{aligned} \right. $$ The derivative has two minima (just expand and simplify, noting that whenever you know both sides are nonnegative, you can square both sides), one of which is the true minimum, for which $p = 4 ( \cos 15^\circ ) ( \sin 15^\circ ) = 2 \sin 30^\circ = 1$, $x = \sqrt{3/4} \approx 0.866$, $y = \sqrt{3/4}(2 - \sqrt{3}) \approx 0.232$
Without symmetries, we can define $E = (x_1, 0)$, $C = (\cos \theta, \sin\theta)$, and $D = (x_2, x_2 \tan 30^\circ) = (x_2, x_2 / \sqrt{3})$. The perimeter length is then $$\begin{aligned} p & = \lVert ED \rVert + \lVert DC \rVert + \lVert CE \rVert \\ ~ & = ~ \sqrt{ (x_2 - x_1)^2 + \left(\frac{x_2}{\sqrt{3}} - 0\right)^2 } \\ ~ & + ~ \sqrt{ (\cos\theta - x_2)^2 + \left(\sin\theta - \frac{x_2}{\sqrt{3}}\right)^2 } \\ ~ & + ~ \sqrt{ (x_1 - \cos\theta)^2 + (0 - \sin\theta)^2} \\ \end{aligned}$$ i.e. $$\begin{aligned} p & = ~ \sqrt{ \frac{4}{3} x_2^2 + x_1^2 - 2 x_1 x_2 } \\ ~ & + ~ \sqrt{ \frac{4}{3} x_2^2 - 2 x_2 \left( \frac{\sin\theta}{\sqrt{3}} + \cos\theta \right) + 1 } \\ ~ & + ~ \sqrt{ x_1^2 - 2 x_1 \cos\theta + 1 } \\ \end{aligned}$$ I don't know of a good trivariate minimization for this, but brute force numerical evaluation does trivially find the solution $p = 1$, $\theta = 15^\circ$, $x_1 \approx 0.89657548$, $x_2 \approx 0.77645714$.
The symmetry is not obvious here from the numbers, but $E = (0.89657548, 0)$ and $D = (0.77645714, 0.44828774)$ are symmetrically positioned around $OC$, $O = (0,0)$, $C = (0.96592583, 0.25881905)$.
Personally, I find this a good example of the situation when finding the easiest way to solve a problem is the key.
