3
$\begingroup$

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,

Triangle

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: SQUARE

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 $\sin 45^\circ=\dfrac{1}{\sqrt 2}$?
$\endgroup$
4
$\begingroup$

You can prove that the two isosceles triangles are congruent by SSS, and hence the angles adding up to 90 are equal.

$\endgroup$
  • $\begingroup$ Thanks for the answer. It was quit simple. $\endgroup$ – user103816 Jun 19 '15 at 11:00
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I don't get it. Perpendicular bisector of what? $\endgroup$ – user103816 Jun 19 '15 at 11:16
  • $\begingroup$ @user103816 Sorry, I'm tired and horrible at English. The angle bisector. $\endgroup$ – SBareS Jun 19 '15 at 11:17
  • $\begingroup$ Are there some other geometric proofs for it? $\endgroup$ – user103816 Jun 23 '15 at 13:56
  • $\begingroup$ @user103816 Of course. You can do something very similar to what David Quinn did. $\endgroup$ – SBareS Jun 23 '15 at 17:49
  • $\begingroup$ I mean some thing else than these methods. $\endgroup$ – user103816 Jun 24 '15 at 3:07
1
$\begingroup$

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 $$\angle{ADB}=\angle{CDB}.$$

$\endgroup$
  • 1
    $\begingroup$ Thanks for the answer. It was quit simple. $\endgroup$ – user103816 Jun 19 '15 at 11:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.