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# Heron's Formula for Triangular Area

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THEOREM

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Figure 6
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For a triangle having sides of length *a*, *b*, and *c* and area *K*, we have

*K* = sqrt[ *s*(*s - a*)(*s - b*)(*s - c*) ],
where *s* = ½(*a* + *b* + *c*) is the triangle's semi-perimeter.

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PROOF

Let *ABC* be an arbitrary triangle. Also, let the side *AB* be at least as long as the other two sides (Figure 6).
Because the proof of Heron's Formula is "circuitous" and long, we'll divide the proof into three main parts.

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Part A

Let *O* be the center of the inscribed circle. Let *r* be the radius of this circle (Figure 7). As we can see, *OD* = *OE* = *OF* = *r*.
Now, applying the usual formula for the area of triangles, we get:

Area(*AOB*) = ½(base)(height) = ½(*AB*)(*OD*) = ½*cr*
Area(*BOC*) = ½(base)(height) = ½(*BC*)(*OE*) = ½*ar*

Area(*COA*) = ½(base)(height) = ½(*AC*)(*OF*) = ½*br*

So,
*K* = Area(*ABC*) = Area(*AOB*) + Area(*BOC*) + Area(*COA*)
By substitution,
*K* = ½*cr* + ½*ar* + ½*br* = ½*r*(*a* + *b* + *c*) = *rs*

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Figure 7
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Part B

Now, using Proposition 1, we can see that *ABC* is now made of three pairs of congruent triangles, namely
*AOD* is congruent to *AOF*
*BOD* is congruent to *BOE*
and *COE* is congruent to *COF*
These congruences follow by AAS. Since corresponding parts of congruent triangles are congruent, we have
*AD* = *AF*, *BD* = *BE*, and *CE* = *CF*
while <*AOD* = <*AOF*, <*BOD* = <*BOE*, and <*COE* = <*COF*.
Now, extend the triangle's base *AB* to a point *G*, such that *AG* = *CE*. Now,

*BG* = *BD* + *AD* + *AG* = *BD* + *AD* + *CE* by construction

= ½(2*BD* + 2*AD* + 2*CE*)
= ½[(*BD* + *BE*) + (*AD* + *AF*) + (*CE* + *CF*)] by congruence
= ½[(*BD* + *AD*) + (*BE* + *CE*) + (*AF* + *CF*)]
= ½[*AB* + *BC* + *AC*]
= ½(*c* + *a* + *b*) = *s*
Now, since *BG* = *s*,

*s* - *c* = *BG* - *AB* = *AG*

*s* - *b* = *BG* - *AC*

= (*BD* + *AD* + *AG*) - (*CF* + *AF*)
= (*BD* + *AD* + *CE*) - (*CE* + *AD*) = *BD* since *AF = AD*, *CF* = *CE* and *AG* = *CE*.
Likewise,
*s* - *a* = *BG* - *BC*
= (*BD* + *AD* + *AG*) - (*BE* + *CE*)
= (*BD* + *AD* + *CE*) - (*BD* + *CE*) = *AD* since *BD* = *BE* and *AG* = *CE*.
These equalities just help us to visualize the segments *s* - *a*, *s* - *b*, and *s* - *c*.

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Part C

Now, construct *OL* such that it's perpendicular to *OB*, intersecting *AB* at *K* (Figure 8).

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Figure 8
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Next, construct *AM* such that it's perpendicular to *AB*, intersecting *OL* at *H* (Figure 9).

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Figure 9
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Next, connect *BH* (Figure 10).

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Figure 10
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By Proposition 4, the quadrilateral AHBO is a cyclic quadrilateral.

By Proposition 5, we know that <*AHB* + <*AOB* = 180 degrees. **(*)**
Please don't be discouraged with the following algebra. These can be visualized using Figure 10.

Now, let's focus on the angles about *O*.

From the congruences in part B, we see that
<*FOA* = <*DOA*, <*DOB* = <*EOB*, and <*EOC* = <*FOC*. **(**)**
Obviously,
<*FOA* + <*DOA* + <*DOB* + <*EOB* + <*EOC* + <*FOC* = 360 degrees.
By **(**)**,
2<*DOA* + 2<*DOB* + 2<*FOC* = 360 degrees,
or
<*DOA* + <*DOB* + <*FOC* = 180 degrees. **(***)**
Obviously, <*DOA* + <*DOB* = <*AOB*. **(****)**

Substituting **(****)** into **(***)**, we see that <*AOB* + <*FOC* = 180 degrees. **(*****)**
From **(*)** and **(*****)**, <*AHB* = <*FOC*.
Now, we can see that triangle *COF* is similar to triangle *BHA*.

By similar triangles,
*AB* / *AH* = *CF* / *OF* = *AG* / *r*
since *CF* = *AG* and *OF* = *r*.
Equivalently,

*AB* / *AG* = *AH* / *r* **(V*)**
Similarly, triangle *KAH* is similar to *KDO*. This is because <*KAH* and <*KDO* are both 90 degrees, while <*AKH* = <*DKO*.

Again by similar triangles,
*AH* / *AK* = *OD* / *KD* = *r* / *KD*
Rearranging yields:
*AH* / *r* = *AK* / *KD*
Combining this with **(V*)**, we get:

*AB* / *AG* = *AK* / *KD* **(V**)**
since both sides are equal to *AH* / *r*.
Now, look at triangle *BOK*. By Proposition 2, we see that triangle *ODB* and triangle *KDO* are similar.

By similar triangles, *KD / r* = *r / BD*
That is,
(*KD*)(*BD*) = *r*^{2} **(V***)**
Now, add 1 to both sides of **(V**)**:

[*AB* / *AG*] + 1 = [*AK / KD*] + 1
Equivalently,
[*AB* + *AG*] / *AG* = [*AK* + *KD*] / *KD*
or
*BG / AG = AD / KD*
Multiplying this equation on the left by *BG/BG* and the right by *BD/BD* yields:

[(*BG*)(*BG*)] / [(*AG*)(*BG*)] = [(*AD*)(*BD*)] / [(*KD*)(*BD*)]
Substituting **(V***)** into this and simplifying, we get
[(*BG*)^{2}] / [(*AG*)(*BG*)] = [(*AD*)(*BD*)] / [*r*^{2}]
Cross-multiply:
*r*^{2}(*BG*)^{2} = (*AG*)(*BG*)(*AD*)(*BD*)
Substituting the segments we derived in part B into the equation above, we get:
*r*^{2}*s*^{2} = (*s - c*)(*s*)(*s - a*)(*s - b*) = *s*(*s - a*)(*s - b*)(*s - c*)
Therefore,
*rs* = sqrt[ *s*(*s - a*)(*s - b*)(*s - c*) ]
From Part A, the area of triangle *ABC = K = rs*.

This gives Heron's Formula:

*K* = sqrt[ *s*(*s - a*)(*s - b*)(*s - c*) ]