175. A Geometric Proof (without trig.) of Heron's Formula.

by Virgil Nicula, Nov 22, 2010, 12:57 PM

Heron's Formula. Prove without trigonometry that the area $S$ of $\triangle ABC$ is given by the relation $S=\sqrt {s(s-a)(s-b)(s-c)}$ , where $2s=a+b+c$ .


Method 1 (Classic). Suppose w.l.o.g. $a\ge b\ge c\ (*)$ . Let $H\in BC$ for which $AH\perp BC$ . Thus, $(*)\implies H\in (BC)$ . Apply generalized Pythagora's

theorem : $b^2=a^2+c^2-2a\cdot HB$ $\implies$ $2a\cdot HB=a^2+c^2-b^2$ $\implies$ $\underline{16S}^2=4a^2h_a^2=$ $4a^2\cdot\left(c^2-HB^2\right)=$ $4a^2c^2-\left(a^2+c^2-b^2\right)^2=$

$\left[(a+c)^2-b^2\right]\left[b^2-(a-c)^2\right]=$ $\underline{16s(s-a)(s-b)(s-c)}$ $\implies$ $S=\sqrt {s(s-a)(s-b)(s-c)}$ .


Method 2 (Shannon Umberger). Denote the incircle $w=C(I,r)$ and $D\in BC\cap w$ , $E\in AC\cap w$ . Is well-known that $S=sr$ , $AE=s-a$ , $DB=s-b$ ,

$DC=s-c$ . Define the points $J\in BC$ for which $IJ\perp IB$ and $H\in IJ$ for which $HC\perp BC$ . Observe that the quadrilateral $BICH$ is cyclically and

$m\left(\widehat{BHC}\right)=180^{\circ}-m\left(\widehat{BIC}\right)=$ $180^{\circ}-\left(90^{\circ}+\frac A2\right)=$ $90^{\circ}-\frac A2=$ $m\left(\widehat{AIE}\right)$ , i.e. $\widehat {BHC}\equiv\widehat {AIE}$ $\iff$ $\triangle BHC\sim \triangle AIE$ .Therefore,

$\frac {BC}{AE}=\frac {HC}{IE}=\frac {HC}{ID}=\frac {JC}{JD}$ $\implies$ $\frac {JC}{a}=\frac {JD}{s-a}=\frac {s-c}{s}$ $\implies$ $JD=\frac {(s-a)(s-c)}{s}$ . Thus, $r^2=ID^2=DB\cdot DJ=$ $(s-b)\cdot \frac {(s-a)(s-c)}{s}$ $\implies$

$\boxed{\ sr^2=(s-a)(s-b)(s-c)\ }$ . In conclusion, $S^2=s\cdot sr^2=s(s-a)(s-b)(s-c)$ $\implies$ $S=\sqrt {s(s-a)(s-b)(s-c)}$ .


Method 3 (Virgil Nicula). Apply Stewart's relation to $ID$ in $\triangle BIC\ :\ IB^2\cdot DC+IC^2\cdot DB=ID^2\cdot BC+DB\cdot DC\cdot BC$ . Using well-known relation

$\boxed{IA^2=\frac {bc(s-a)}{s}}$ a.s.o. obtain that $\frac {ac(s-b)}{s}\cdot (s-c)+\frac {ab(s-c)}{s}\cdot (s-b)=ar^2+a(s-b)(s-c)$ $\iff$ $(s-b)(s-c)(c+b-s)=sr^2$ $\iff$

$\boxed{\ (s-a)(s-b)(s-c)=sr^2\ }$ because $b+c-s=s-a$ . Thus, $S^2=s\cdot sr^2=s(s-a)(s-b)(s-c)$ $\implies$ $S=\sqrt {s(s-a)(s-b)(s-c)}$ .

Remark. Must prove synthetically (without trigonometry) the well-known relation $IA^2=\frac {bc(s-a)}{s}$ a.s.o.

Proof. Denote $L\in AI\cap BC$ and the second intersection $S$ of $AI$ with the circumcircle of $\triangle ABC$ . Since $\triangle ABS\sim\triangle ALC$

obtain that $AL\cdot AS=bc$ . Observe that $\frac {AI}{AI_a}=\frac {s-a}{s}=\frac {r}{r_a}=\frac {LI}{LI_a}$ , i.e. the division $(A,I,L,I_a)$ is harmonically. Since $S$ is

the midpoint of $[II_a]$ obtain that $AI\cdot AI_a=AL\cdot AS$ , i.e. $AI\cdot AI_a=bc$ . In conclusion, $IA^2=\frac {AI}{AI_a}\cdot bc=$ $\frac {bc(s-a)}{s}$ .


Method 4 (trigonometric). From the sum and subtraction of the relations $\begin{array}{c} s(s-a)+(s-b)(s-c)=bc\\\\ s(s-a)-(s-b)(s-c)=bc\cdot\cos A\end{array}$ obtain that $\begin{array}{c} \frac {1+\cos A}{2}=\frac {s(s-a)}{bc}\\\\ \frac {1-\cos A}{2}=\frac {(s-b)(s-c)}{bc}\end{array}$ .

From the relations $\begin{array}{c} \cos^2\frac A2=\frac {1+\cos A}{2}\\\\ \sin^2\frac A2=\frac {1-\cos A}{2}\end{array}$ obtain that $\boxed{\begin{array}{c} \cos\frac A2=\sqrt {\frac {s(s-a)}{bc}}\\\\ \sin\frac A2=\sqrt {\frac {(s-b)(s-c)}{bc}}\end{array}}$ and $\boxed{\tan\frac A2=\sqrt {\frac {(s-b)(s-c)}{s(s-a)}}}$ . Since $\tan\frac A2=\frac {IE}{AE}=$ $\frac {r}{s-a}\implies $ $\boxed{r=(s-a)\cdot\tan\frac A2}\implies$ $S=sr=s(s-a)\tan\frac A2=s(s-a)\cdot \sqrt {\frac {(s-b)(s-c)}{s(s-a)}}$ $\implies S=\sqrt {s(s-a)(s-b)(s-c)}$ .

Remark. $S=\frac {bc\cdot \sin A}{2}=$ $bc\cdot\sin\frac A2\cdot\cos\frac A2=$ $bc\cdot\sqrt {\frac {(s-b)(s-c)}{bc}}\cdot\sqrt {\frac {s(s-a)}{bc}}$ $\implies$ $S=\sqrt {s(s-a)(s-b)(s-c)}$ .
This post has been edited 53 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:05 PM

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13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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    by ryanbear, May 9, 2023, 6:10 AM

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    by OlympusHero, Sep 14, 2022, 4:44 AM

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    by tigerzhang, Aug 1, 2021, 12:02 AM

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    by the_mathmagician, Jan 17, 2021, 7:43 PM

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    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

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