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My brother was told to find the area of a right-angled triangle. He knew that if he made a copy of the right-angle triangle so that there were two, joining the longest side of each triangle together, he would form a square or a rectangle. Because of this he knew that the area of the right-angled triangle was half the area of the square/rectangle.

Let:

$p = \text{area of right-angled triangle;}$

$q = \text{area of rectangle;}$

$l = \text{length of the rectangle; and}$

$w = \text{height (altitude) of the rectangle.}$

Given this information, my brother was able to find out that: $$p = \left\{\frac{q}{2} : q = lw\right\}$$

He then asked me, why the area of any triangle uses the same formula, because you can't always make two copies of a triangle, join the longest sides together, and then make a square/rectangle. It sometimes makes a parallelogram depending on your triangle!

I replied that for any four-sided shape, the area will always equal its length multiplied by its width. It does not have to be a square or rectangle in order for it to work. The formula for finding the area of a square is the same for a rectangle, but is also the same for a diamond and the same for a parallelogram as well.

Then my brother said to me that because the same formula applied to all four-sided shapes, he did not have to join two copies of the triangle together. He drew an equilateral triangle $\triangle$ and then fitted a square over it $\square$ such that the bottom length of the square was the base of the triangle. enter image description here Then my brother said to me;

"If we label the sides of the square $x$ since they are equal to each other, and the area of the triangle $y$, how can you prove that $y = \frac{x^2}{2}$?"

Well, I know that the two blue seperate areas inside the square are not covered by the triangle, thus if we add them together, they will equal the area of the triangle. Let's call these particular areas $z_1$ and $z_2$. Both $z_1$ and $z_2$ are also triangles, so for some value $a$ and some value $b$, $\frac{ab}{2}$ is equal to the area of $z_1$ and $z_2$, such that $a$ is equal to each long side length of $z_1$ and $z_2$, and $b$ is equal to each short side length of $z_1$ and $z_2$. (Here we exclude the longest side of the triangles, since each triangle has three sides but we are only including two in this case.)

I knew that I could use the same method my brother used of finding the value of $p$ (the area of the right-angled triangle he was looking for), meaning that I could actually split the square in half so it is combined with two rectangles $r_1$ and $r_2$, drawing a line straight down from the top corner of the triangle across its centre. The width of $r_1$ is $m$ and the width of $r_2$ is $n$.

I know that: $$m + n = \left\{x : x = \text{length of the square, namely the length of the base of the triangle}\right\}$$ And I also know that: $$y = z_1 + z_2 = \frac{mx}{2} + \frac{nx}{2} = \frac{mx + nx}{2} = \left\{\frac{x(m + n)}{2} : y = \text{area of triangle}\right\}$$ And since we already clarified that $m + n = x$, this proves that $y = \frac{x^2}{2}$ and thus if we covered a rectangle or parallelogram etc over the triangle instead, labelling the length $l$ and the width $w$, the area of the triangle would equal $\frac{lw}{2}$.

I was actually very proud that I had done this, because it took me about $15$-$20$ minutes to prove this, and I showed my science teacher to see if I had done it correctly. (I didn't have maths, but I did have science and my science teacher is fairly good at maths. Note, this is yesterday I am talking about.) He said that I did it correctly, but now he has another challenge for me (assuming that I have spare time to complete it after I did the science homework, which I do), however, when I was doing the challenge last night, I didn't know where to begin:

Challenge:

Prove that the area of any triangle can be obtained using Heron's Fomula: $$\begin{align} A_t = \sqrt{s(s - a)(s - b)(s - c)} \ : A_t &= \text{Area of triangle} \\ s &= \frac{a + b + c}{2} \\ (a, b, c) &= \text{Side lengths of triangle} \end{align}$$

Extra: If you have an equilateral triangle with a side-length $x$, drawing a line $y$ from the top corner down across the centre, how long would $y$ be such that $(x, y)\in \mathbb{R}$?

For the Extra part, according to the given information, $y$ must include $x$, and according to the Pythagorean Theorem $a^2 + b^2 = c^2$, I may be able to find $y$.

Since we are splitting the equilateral triangle in half, we would have two right-angled triangles, where their longest side is $x$, their second longest side is $y$, and their shortest side is of course $\frac{x}{2}$.

$$\begin{align} \therefore y &= \sqrt{x - \bigg(\frac{x}{2}\bigg)^2} \\ &= \sqrt{x - \bigg(\frac{x^2}{2^2}\bigg)} \\ &= \sqrt{x - \bigg(\frac{x^2}{4}\bigg)} \\ &= \sqrt{\frac{4x - x^2}{4}} \\ &= \sqrt{\frac{x(4 - x)}{4}} \\ &= \frac{\sqrt{x(4 - x)}}{2} \\ \end{align}$$ But: $$(x, y)\in \mathbb{R} \implies x\in \mathbb{R} \implies x \leqslant 4 \ \lor \ x \geqslant 4$$ And: $$y\in \mathbb{C}\notin \mathbb{R} \iff x > 4$$ So this becomes a problem, and for the part concerning Heron's Formula, I am not familiar with it and only heard about it when my science teacher told me about it. Could you please help me with this and show me what to do? I would appreciate it if you gave me an example of how this formula works and how this Heron person discovered it. (And trust me, I won't steal all the credit when I show my science teacher the proof.)

Thank you in advance.

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    $\begingroup$ galileoandeinstein.physics.virginia.edu/more_stuff/Heron.html $\endgroup$ – Khosrotash Sep 9 '17 at 11:11
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    $\begingroup$ @Khosrotash thank you! I also found another link here: $$\longrightarrow \text{se16.info/hgb/triangle.htm}$$ and here: $$\longrightarrow \text{math.stackexchange.com/questions/576831/…}$$ explaining the proof. But none of the links explain where I went wrong for the Extra part. $\endgroup$ – Mr Pie Sep 9 '17 at 11:17
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    $\begingroup$ Whoops something happened to the comment and I can't fix it up because it's too late, but I'm sure you know what I mean $\endgroup$ – Mr Pie Sep 9 '17 at 11:23
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    $\begingroup$ In your extra part you forgot to square $x$. $y=\frac{\sqrt{3}}{2}x$ $\endgroup$ – Vasya Sep 9 '17 at 11:40
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    $\begingroup$ So $$y = \frac{x\sqrt{3}}{2}$$ Ok thanks for that :) $\endgroup$ – Mr Pie Sep 9 '17 at 12:40
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enter image description here

Since $$ \begin{align} a^2&=c_a^2+h^2\\ b^2&=c_b^2+h^2 \end{align} $$ we get $$ \begin{align} c_a+c_b&=c\\ c_a-c_b&=\frac{a^2-b^2}c \end{align} $$ therefore, $$ c_a=\frac{a^2-b^2+c^2}{2c} $$ The square of the area is $$ \begin{align} A^2 &=\frac14c^2h^2\\ &=\frac14c^2\left(a^2-c_a^2\right)\\ &=\frac14c^2\left(a-c_a\right)\left(a+c_a\right)\\ &=\frac14c^2\left(a-\frac{a^2-b^2+c^2}{2c}\right)\left(a+\frac{a^2-b^2+c^2}{2c}\right)\\ &=\frac1{16}\left(2ac-a^2-c^2+b^2\right)\left(2ac+a^2+c^2-b^2\right)\\[3pt] &=\frac1{16}\left(b^2-(a-c)^2\right)\left((a+c)^2-b^2\right)\\[3pt] &=\frac{(b-a+c)(b+a-c)(a+c-b)(a+c+b)}{16}\\[6pt] &=(s-a)(s-c)(s-b)s \end{align} $$

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  • $\begingroup$ This is actually a much better answer than Michael’s, purely because of the picture even though I understand anyway. Thank you very much nonetheless! $$(+1) \ \ \color{green}{\checkmark}$$ $\endgroup$ – Mr Pie Dec 21 '17 at 1:57
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    $\begingroup$ This is a special case of Brahmagupta's Theorem of which I provide a proof in this answer. $\endgroup$ – robjohn Dec 21 '17 at 2:37
  • $\begingroup$ I think you should put more explanation on those steps. for example, why ca-cb is (a^2-b^2)/c. I know why. You factor a^2-b^2 factorization. But most won't. Also how do you get good mathtex editor? $\endgroup$ – user4951 May 11 '18 at 15:05
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enter image description here

This is from an article I remember reading in an AMS journal.

Let $D$ be the point of intersection of the internal bisectors of $\angle A$, $\angle B$, and $\angle C$. Let $\alpha$, $\beta$, and $\gamma$ be the measures of the corresponding bisected angles. Note $$\alpha + \beta + \gamma = \frac 12 \pi.$$ Let $r$ be the common distance from the point $D$ to the sides $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$. Define $$s = \frac 12(a+b+c).$$ Note that the feet of the perpendiculars drawn from point $D$ to the sides $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$ create segments with the indicated lengths. Let $u,v,$ and $w$ be the respective lengths of the segments drawn from point $D$ to points $A$, $B$, and $C$. Then

\begin{align} (s-a)+ir = ue^{i \alpha}\\ (s-b)+ir = ve^{i \beta}\\ (s-c)+ir = we^{i \gamma}\\ \end{align}

So $(s-a+ir)(s-b+ir)(s-c+ir)=uvwe^{\frac 12 i\pi}=iuvw$. Hence the real part of $(s-a+ir)(s-b+ir)(s-c+ir)$ is equal to $0$.

We compute \begin{align} \Re[(s-a+ir)(s-b+ir)(s-c+ir)] &= 0 \\ (s-a)(s-b)(s-c)-r^2[(s-a)+(s-b)+(s-c)] &= 0 \\ sr^2 &= (s-a)(s-b)(s-c) \\ r &= \sqrt{\frac{(s-a)(s-b)(s-c)}{s}} \end{align}

Hence the area of $\triangle ABC$ is

$$ \frac 12r(a+b+c) = rs = \sqrt{s(s-a)(s-b)(s-c)} $$

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  • $\begingroup$ I understand everything apart from the second half, but that is ok. I acknowledge the effort you put into formatting this post. $(+1)$ $\endgroup$ – Mr Pie Mar 24 '18 at 13:50
  • $\begingroup$ Dude what the hell is point P? You mean point D? $\endgroup$ – user4951 May 11 '18 at 15:09
  • $\begingroup$ @J.Chang - Thanks. I fixed it. $\endgroup$ – steven gregory May 11 '18 at 16:59
  • $\begingroup$ Very great. Why does distance between a and the r are (s-a). I didn't get that part. Looks correct. But why? $\endgroup$ – user4951 May 11 '18 at 17:27
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    $\begingroup$ @J.Chang - Not that I know of. I would love to see other proofs where complex numbers show up out of nowhere and turn out to be helpful. $\endgroup$ – steven gregory May 18 '18 at 15:19
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Let $AB=c$, $AC=b$, $BC=a$, $a=\max\{a,b,c\}$ and $AD$ be an altitude of $\Delta ABC$.

Thus, $D$ placed between $B$ and $C$.

Hence, by the Pythagoras's theorem $$AH^2=c^2-BD^2=b^2-(a-BD)^2.$$

Thus, $$c^2-BD^2=b^2-a^2+2aBD-BD^2,$$ which gives $$BD=\frac{a^2+c^2-b^2}{2a}$$ and from here $$AD=\sqrt{c^2-BD^2}=\sqrt{c^2-\left(\frac{a^2+c^2-b^2}{2a}\right)^2}=$$ $$=\frac{1}{2a}\sqrt{4a^2c^2-(a^2+c^2-b^2)^2}=$$ $$=\frac{1}{2a}\sqrt{2ac-a^2-c^2+b^2)(2ac+a^2+c^2-b^2)}=$$ $$=\frac{1}{2a}\sqrt{(b^2-(a-c)^2)((a+c)^2-b^2)}=$$ $$=\frac{1}{2a}\sqrt{(a+b-c)(b+c-a)(a+b+c)(a+c-b)}=$$ $$=\frac{1}{2a}\sqrt{(a+b+c-2c)(b+c+a-2a)(a+b+c)(a+c+b-2b)}=$$ $$=\frac{1}{2a}\sqrt{(2s-2c)(2s-2a)(2s)(2s-2b)}=$$ $$=\frac{2}{a}\sqrt{s(s-a)(s-b)(s-c)}.$$ Id est, $$S_{\Delta ABC}=\frac{1}{2}\cdot a\cdot\frac{2}{a}\sqrt{s(s-a)(s-b)(s-c)}=$$ $$=\sqrt{s(s-a)(s-b)(s-c)}.$$ and we are done!

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  • $\begingroup$ I did not know what max{${a, b, c}$} meant but I did a little bit of research and correct me if I am wrong, but I think if this equals $a$ then that implies that $a > b\land c$. The rest, I understand :) Thank you so much!! You actually gave a better explanation in your answer than the links provided! $$(+1) \ \ \checkmark$$ and we are done! $\endgroup$ – Mr Pie Sep 9 '17 at 12:38
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    $\begingroup$ @user477343 I want that $D$ will be between $B$ and $C$ for which we can assume that $a\geq b$ and $a\geq c$. We can assume this think because there is a symmetry in our reasoning. $\endgroup$ – Michael Rozenberg Sep 9 '17 at 12:43

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