Trigonometry/Triangle Geometry

Welcome to the Lesson of Trilaterometry

Part of the School of Olympiads

Trilaterometry or Triangle Geometry refers to the aspect of plane geometry which deals with triangles, which are the the most basic polygon with only 3 (minimum possible) sides. Its importance can be justified by the fact that all polygons can cut into triangles with all its vertices being vertices of original polygon.

Theorems

Area of a Triangle

The area of a triangle can be demonstrated as half of the area of a paralellogram which has the same base length and height.

Calculating the area of a triangle is an elementary problem encountered often in many different situations. The best known and simplest formula is:

$A={\frac {1}{2}}bh$

where $A$ is area, $b$ is the length of the base of the triangle, and $h$ is the height or altitude of the triangle. The term 'base' denotes any side, and 'height' denotes the length of a perpendicular from the point opposite the side onto the side itself.

Although simple, this formula is only useful if the height can be readily found. For example, the surveyor of a triangular field measures the length of each side, and can find the area from his results without having to construct a 'height' with the Heron's Formula.

The shape of the triangle is determined by the lengths of the sides alone. Therefore the area S also can be derived from the lengths of the sides. By Heron's formula:

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

where s = ½ (a + b + c) is the semiperimeter, or half of the triangle's perimeter.

Three equivalent ways of writing Heron's formula are

$S={\frac {1}{4}}{\sqrt {(a^{2}+b^{2}+c^{2})^{2}-2(a^{4}+b^{4}+c^{4})}}$

$S={\frac {1}{4}}{\sqrt {2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{4}+b^{4}+c^{4})}}$

$S={\frac {1}{4}}{\sqrt {(a+b-c)(a-b+c)(-a+b+c)(a+b+c)}}.$

Heron's formula is a special case of Brahmagupta's Theorem.

Intercept Theorem

The intercept theorem is an important theorem in elementary geometry about the ratios of various line segments, that are created if 2 intersecting lines are intercepted by a pair of parallels. It is equivalent to the theorem about ratios in similar triangles. Traditionally it is attributed to Greek mathematician Thales, which is the reason why it is named theorem of Thales in some languages.

Theorem and Proof:

Due to heights of equal length ($CA\parallel BD$ ) we have $|\triangle CDA|=|\triangle CBA|$ and therefore $|\triangle SCB|=|\triangle SDA|$. This yields ${\frac {|\triangle SCA|}{|\triangle CDA|}}={\frac {|\triangle SCA|}{|\triangle CBA|}}$ and ${\frac {|\triangle SCA|}{|\triangle SDA|}}={\frac {|\triangle SCA|}{|\triangle SCB|}}$ Plugging in the actual formula for triangle areas (${\frac {baseline\cdot height}{2}}$) transforms that into ${\frac {|SC||AF|}{|CD||AF|}}={\frac {|SA||EC|}{|AB||EC|}}$ and ${\frac {|SC||AF|}{|SD||AF|}}={\frac {|SA||EC|}{|SB||EC|}}$ Cancelling the common factors results in: (a) $\,{\frac {|SC|}{|CD|}}={\frac {|SA|}{|AB|}}$ and (b) $\,{\frac {|SC|}{|SD|}}={\frac {|SA|}{|SB|}}$ Now use (b) to replace $|SA|$ and $|SC|$ in (a): ${\frac {\frac {|SA||SD|}{|SB|}}{|CD|}}={\frac {\frac {|SB||SC|}{|SD|}}{|AB|}}$ Using (b) again this simplifies to: (c) $\,{\frac {|SD|}{|CD|}}={\frac {|SB|}{|AB|}}$ $\,\square$

Pythagorean Theorem

The Pythagorean theorem: The sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c).

In mathematics, the Pythagorean theorem (American English) or Pythagoras' theorem (British English) is a relation in Euclidean geometry among the three sides of a right triangle. The theorem is usually written as an equation:

$a^{2}+b^{2}=c^{2}\!\,$

where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides. In words:

The square of the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides.

The Pythagorean theorem is named after the Greek mathematician Pythagoras, who by tradition is credited with its discovery and proof, although it is often argued that knowledge of the theory predates him. (There is much evidence that both Babylonians and Indian understood the principle).

Proof 1:

Proof using similar triangles

Like most of the proofs of the Pythagorean theorem, this one is based on the proportionality of the sides of two similar triangles.

Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. We draw the altitude from point C, and call H its intersection with the side AB. The new triangle ACH is similar to our triangle ABC, because they both have a right angle (by definition of the altitude), and they share the angle at A, meaning that the third angle will be the same in both triangles as well. By a similar reasoning, the triangle CBH is also similar to ABC. The similarities lead to the two ratios..:

As

$BC=a,AC=b,{\text{ and }}AB=c,\!$

so

${\frac {a}{c}}={\frac {HB}{a}}{\mbox{ and }}{\frac {b}{c}}={\frac {AH}{b}}.\,$

These can be written as

$a^{2}=c\times HB{\mbox{ and }}b^{2}=c\times AH.\,$

Summing these two equalities, we obtain

$a^{2}+b^{2}=c\times HB+c\times AH=c\times (HB+AH)=c^{2}.\,\!$

In other words, the Pythagorean theorem:

$a^{2}+b^{2}=c^{2}.\,\!$

Proof 2 (as given in Baudhayana SulbaSutra):

Algebraic proof: A square created by aligning four right angle triangles and a large square

Looking at the illustration which is a large square with identical right triangles in its corners, the area of each of these four triangles is given by an angle corresponding with the side of length C.

${\frac {1}{2}}AB.$

The A-side angle and B-side angle of each of these triangles are complementary angles, so each of the angles of the blue area in the middle is a right angle, making this area a square with side length C. The area of this square is C². Thus the area of everything together is given by:

$4\left({\frac {1}{2}}AB\right)+C^{2}.$

However, as the large square has sides of length A + B, we can also calculate its area as (A + B)², which expands to A² + 2AB + B².

$A^{2}+2AB+B^{2}=4\left({\frac {1}{2}}AB\right)+C^{2}.\,\!$

(Distribution of the 4) $A^{2}+2AB+B^{2}=2AB+C^{2}\,\!$

(Subtraction of 2AB) $A^{2}+B^{2}=C^{2}\,\!$

Angle Bisector Theorem

In this diagram, BD:DC = AB:AC.

In geometry, the angle bisector theorem relates the length of the side opposite one angle of a triangle to the lengths of the other two sides of the triangle.

Consider a triangle ABC. Let the angle bisector of angle A intersect side BC at a point D. The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC

${\frac {|AB|}{|AC|}}={\frac {|BD|}{|DC|}}.$

The generalized angle bisector theorem states that if D lies on BC, then

${\frac {|BD|}{|DC|}}={\frac {|AB|\sin \angle DAB}{|AC|\sin \angle DAC}}.$

This reduces to the previous version if AD is the bisector of BAC.

Proof of generalization:

Let B₁ be the base of altitude in the triangle ABD through B and let C₁ be the base of altitude in the triangle ACD through C. Then,

${\begin{aligned}|BB_{1}|&=|AB|\sin \angle BAD,\\|CC_{1}|&=|AC|\sin \angle CAD.\end{aligned}}$

It is also true that both the angles DB₁B and DC₁C are right, while the angles B₁DB and C₁DC are congruent if D lies on the segment BC and they are identical otherwise, so the triangles DB₁B and DC₁C are similar (AAA), which implies that

${\frac {|BD|}{|CD|}}={\frac {|BB_{1}|}{|CC_{1}|}}={\frac {|AB|\sin \angle BAD}{|AC|\sin \angle CAD}}.$

Q.E.D.

Apollonius' Theorem

In elementary geometry, Apollonius' theorem is a theorem relating several elements in a triangle.

It states that given a triangle ABC, if D is any point on BC such that it divides BC in the ratio n:m (or $mBD=nDC$), then

$mAB^{2}+nAC^{2}=mBD^{2}+nDC^{2}+(m+n)AD^{2}\,$

Special cases of the theorem:

• When $m=n(=1)$, that is, AD is the median falling on BC, the theorem reduces to

$AB^{2}+AC^{2}=BD^{2}+DC^{2}+2AD^{2}.\,$

• When in addition AB = AC, that is, the triangle is isosceles, the theorem reduces to the Pythagorean theorem,

$AD^{2}+BD^{2}=AB^{2}(=AC^{2}).\,$

In simpler words, in any triangle $ABC\,$, if $AD\,$ is a median, then $AB^{2}+AC^{2}=2(AD^{2}+BD^{2})\,\!$

Ceva's Theorem

Ceva's theorem:the three lines are concurrent at a point O inside ABC

In geometry, Ceva's theorem is a theorem about triangles. Given a triangle ABC, let the lines AO, BO and CO be drawn from the vertices to a common point O (not on one of the sides of ABC), to meet opposite sides at D, E and F respectively. (The segments AD, BE, and CF are known as cevians.) Then, using the lengths of segments,

${\frac {AF}{FB}}\cdot {\frac {BD}{DC}}\cdot {\frac {CE}{EA}}=1.$

Resources

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