Get special measurements from the sides

Consider sides $a,b,c$, area $K$ and semiperimeter $s=(a+b+c)/2$. We require a non-degenerate triangle: positive sides and each side smaller than the sum of the other two.

Demonstrate Heron’s formula

From the law of cosines, $\cos A=(b^2+c^2-a^2)/(2bc)$. In $K=bc\sin A/2$, use $\sin^2A=1-\cos^2A$:

$$\begin{aligned} 16K^2&=4b^2c^2-(b^2+c^2-a^2)^2\\ &=((b+c)^2-a^2)(a^2-(b-c)^2)\\ &=(a+b+c)(b+c-a)\\ &\quad\cdot(a+b-c)(a-b+c)\\ &=16s(s-a)(s-b)(s-c). \end{aligned}$$

The area is positive, so:

$$K=\sqrt{s(s-a)(s-b)(s-c)}.$$

Heights

As $K=ah_a/2=bh_b/2=ch_c/2$:

$$h_a=\frac{2K}{a},\qquad h_b=\frac{2K}{b},\qquad h_c=\frac{2K}{c}.$$

Substituting Heron’s expression gives each height as a function of the sides only. $h_a=b\sin C=c\sin B$ and its permutations are also valid.

Medians

The median $m_a$ connects A to the midpoint of BC. Place the base ends at $(-a/2,0)$ and $(a/2,0)$ and A at $(u,v)$. Adding the squared distances to the endpoints gives:

$$b^2+c^2=2(u^2+v^2)+\frac{a^2}{2}.$$

As $m_a^2=u^2+v^2$, it results:

$$m_a=\frac12\sqrt{2b^2+2c^2-a^2}.$$

Swapping the indexes gives $m_b$ and $m_c$.

angle bisectors internal

The angle bisector of A meets BC in D and divides the angle into two equal parts. The areas of ABD and ACD are in the ratio $c:b$, because they have the same height relative to the line BC and can also be calculated with the sine of $A/2$. Logo:

$$BD=\frac{ac}{b+c},\qquad DC=\frac{ab}{b+c}.$$

The sum of the areas calculated by the sides of angle bisector $t_a=AD$ is:

$$K=\frac{(b+c)t_a\sin(A/2)}{2}.$$

Comparing with $K=bc\sin A/2$ and using $\sin A=2\sin(A/2)\cos(A/2)$:

$$t_a=\frac{2bc\cos(A/2)}{b+c} =\frac{2\sqrt{bc\,s(s-a)}}{b+c}.$$

The second form arises from $\cos^2(A/2)=s(s-a)/(bc)$, obtained by the law of cosines. All roots are positive. Swap the letters to get the rest angle bisectors.

Inscribed and circumscribed circles

The incenter is a distance $r$ on each side. Dividing the triangle into the three triangles that have this center as the vertex gives:

$$K=\frac{ar+br+cr}{2}=sr.$$

Like this:

$$r=\frac Ks=\sqrt{\frac{(s-a)(s-b)(s-c)}s}.$$

The tangent drawn from a vertex to the two points of contact has equal lengths. The section associated with vertex A measures $s-a$. In the right triangle formed with incenter:

$$r=(s-a)\tan\frac A2.$$

Using the law of sines and the area expression from the previous class:

$$R=\frac{abc}{4K}.$$

Example: sides 5, 5 and 6

We have $s=8$ and $K=\sqrt{8\cdot3\cdot3\cdot2}=12$. For base $a=6$, with $b=c=5$:

$$h_a=4,\quad m_a=4,\quad t_a=4,\quad r=\frac32,\quad R=\frac{25}{8}.$$

Height, median and angle bisector coincide in this direction because the triangle is isosceles. In general, they are different segments.

In everyday life

  • Area with measuring tape. Measuring only the sides of a piece of land (13, 14 and 15 m), Heron’s formula gives 84 m².
  • Fountain in the garden. The radius of the largest circular fountain that fits in a triangular flowerbed is the inscribed radius, $r = \frac{S}{p}$.
  • Equilibrium. A triangular plate of uniform thickness balances on the tip of a pencil at centroid, the meeting of the medians.
Heron, inscribed ray and circumscribed ray in a triangular terrain.

Exercises with resolution

1. A triangle has sides 13, 14 and 15 cm. Calculate area and inscribed radius.

Check resolution 1

$s=21$ cm and $K=\sqrt{21\cdot8\cdot7\cdot6}=84$ cm². Then $r=84/21=4$ cm.

2. On the same sides, calculate the height and median relative to the side of 14 cm.

Check resolution 2

$h=2\cdot84/14=12$ cm. The median is $m=\tfrac12\sqrt{2\cdot13^2+2\cdot15^2-14^2}=\sqrt{148}=2\sqrt{37}$ cm.

In the interactive book

The same subjects, with all the demonstrations and the complete list of exercises, are in the interactive Trigonometry book:

This class is a shorter introduction to the topic; In the book, the theory is accompanied by solved and proposed exercises from the chapter.

Stage index · ← Previous class · Next class →