How is $\sin 45^\circ=\frac{1}{\sqrt 2}$? I've been reading about the proof of $\sin 45^\circ=\dfrac{1}{\sqrt 2}$ in my book. They did it as following, let $\triangle ABC$ be an isosceles triangle as shown, 

Since the triangle is isosceles with base and perpendicular equal the opposite angle $\angle C$ and $\angle B$ must be equal. Putting $\angle A=90^\circ$ and using $\angle A + \angle B + \angle C = 180^0$, we get $\angle A= \angle B= 45^\circ$. Now using Pythagoras theorem gives $\sin 45^\circ = \dfrac {1}{\sqrt 2}$. I am satisfied with this explanation. But on https://proofwiki.org they used a square to prove this formula as shown: 
In that proof over proofwiki they say that the $\angle A$ is $45^\circ$ because the diagonal $AC$ is the bisector of $\angle A$. They do not use the property of isosceles triangle. My questions are,


*

*Why is the diagonal of a square bisector of the angle from which it originates? How can we prove this mathematically?

*Are there some other geometric proofs of $\sin 45^\circ=\dfrac{1}{\sqrt 2}$?

 A: You can prove that the two isosceles triangles are congruent by SSS, and hence the angles adding up to 90 are equal.
A: The angle bisector is exactly the line of points equidistant from the two lines it bisects. The distance of $C$ to $AB$ is $BC$ and the distance of $C$ to $AD$ is $CD$. Since $ABCD$ is a square, $BC=CD$ and thus $C$ lies on the angle bisector. Now since the angle bisector also goes through $A$, it must be exactly the line $AC$.
Concerning other proofs, here's another, albeit similar, proof: It is clear from the same construction that $\sin{45^\circ} = \cos{45^\circ}$. Now since 
$$\left(\sin\theta\right)^2+\left(\cos\theta\right)^2=1$$
we get:
$$2\left(\sin{45^\circ}\right)^2=1\Rightarrow\sin{45^{\circ}}=\frac{1}{\sqrt2}$$ 
observing that it must be positive. 
A: 
Why is the diagonal of a square bisector of the angle from which it originates? How can we prove this mathematically?

Let $ABCD$ be a square. We know that two triangles $\triangle{ABD},\triangle{CBD}$ are congruent because $AB=CB,AD=CD,\angle{BAD}=\angle{BCD}$. Now we have $$\angle{ADB}=\angle{CDB}.$$
