Law of Cosines (Cosine Rule)
Consider triangle ABC (figure below):
The height "h" splits triangle ABC into two smaller triangles AHB and AHC.
Applying Pythagoras Theorem to triangle AHC:
$b^2=(a-m)^2+h^2$
$b^2=a^2-2am+m^2+h^2$
$b^2=a^2-2am+(m^2+h^2)$ (equation I)
Applying Pythagoras Theorem to triangle AHB:
$m^2+h^2=c^2$ (equation II)
Plugging equation II into I:
$b^2=a^2-2am+(m^2+h^2)$
$b^2=a^2-2am+c^2$ (equation III)
In right triangle AHB, $m=(c)cos(\hat{B})$. Plugging this result into equation III:
$b^2=a^2+c^2-2am$
$b^2=a^2+c^2-2(ac)cos(\hat{B})$
This is the Cosine Rule:
$a^2=b^2+c^2-2(bc)cos(\hat{A})$
$b^2=a^2+c^2-2(ac)cos(\hat{B})$
$c^2=a^2+b^2-2(ab)cos(\hat{C})$
Applying the Cosine Rule to Parallelograms
Consider parallelogram ABCD (figure below), in which the red lines are its diagonals:
Let's apply the Cosine Rule to triangles ABD and ACD:
(I) Triangle ABD: $(DB)^2=(AB)^2+(AD)^2-2(AB)(AD)cos(\hat{A})$
(II) Triangle ACD: $(AC)^2=(DC)^2+(AD)^2-2(DC)(AD)cos(\hat{D})$
Given that ABCD is a parallelogram, $DC=AB$, and $cos(\hat{D})=cos(\hat{180-A})=-cos(\hat{A})$. Plugging these relations into equation II:
Triangle ACD:
$(AC)^2=(AB)^2+(AD)^2-2(AB)(AD)(-cos(\hat{A}))$
$(AC)^2=(AB)^2+(AD)^2+2(AB)(AD)cos(\hat{A})$ (equation III)
Adding equations I and III:
$(AC)^2+(DB)^2=2(AB)^2+2(AD)^2-2(AB)(AD)cos(\hat{A})+2(AB)(AD)cos(\hat{A})$
$(AC)^2+(DB)^2=2(AB)^2+2(AD)^2$, where AC and DB are the diagonals; AB and AD are the sides of the parallelogram.
Law of Sines
Consider triangle ABC (figure below), inscribed in the circle with radius "r" and center "O":
Triangle AOC is isosceles, since it has two sides of equal length, "r". Thus the two base angles of triangle AOC are congruent, "x".
Triangle AOB is also isosceles, since it has two sides of equal length, "r". Thus the two base angles of triangle AOB are also congruent, "y".
The sum of the three internal angles of triangle ABC is equal to 180º. Thus:
$y+y+x+x+O\hat{B}C+O\hat{C}B=180$
(I) $O\hat{B}C+O\hat{C}B=180-2x-2y$
The sum of the three internal angles of triangle BOC is equal to 180º, Thus:
(II) $B\hat{O}C+O\hat{B}C+O\hat{C}B=180$
Plugging I into II:
$B\hat{O}C+180-2x-2y=180$
$B\hat{O}C=2x+2y=2(B\hat{A}C)$
Angle $B\hat{O}C$ is called central angle;
Angle $B\hat{A}C$ is called inscribed angle.
Conclusion: The inscribed angle is half the central angle that subtends the same arc on the circle, regardless of where on the circle the vertex is placed (figure below).
If angle $B\hat{A}C$ measures 90º, angle $B\hat{O}C$ measures 180º. That is, side BC would be the diameter of the circle (2r).
Now let's move point A to position A', so that angle $B\hat{C}A'$ measures 90º:
In such a case, side BA' will be the perimeter of the circle and measure 2r:
$sin(a)=\frac{BC}{2r}$
$\frac{BC}{sin(a)}=2r$
Repeating these steps with the other two angles of triangle ABC, we find the Law of Sines:
$\frac{BC}{sin(a)}=\frac{AC}{sin(b)}=\frac{AB}{sin(c)}=2r$.
Height of Triangles
Consider triangle ABC (figure below), where $h_a$ is the height, as measured from vertex A to side BC.
(I) $h_a=(c)sin(c)$
According to the Law of Cosines:
$b^2=a^2+c^2-2(ac)cos(c)$
$2(ac)cos(c)=a^2-b^2+c^2$
(II) $cos(c)=\frac{a^2-b^2+c^2}{2ac}$
By the Pythagorean Identity: $sin^2(c)+cos^2(c)=1$. Therefore:
(III) $sin(c)=\sqrt{1-cos^2(c)}$
Plugging (II) into (III):
$sin(c)=\sqrt{1-(\frac{a^2-b^2+c^2}{2ac})^2}$
$sin(c)=\sqrt{(1+\frac{a^2-b^2+c^2}{2ac}).(1-\frac{a^2-b^2+c^2}{2ac})}$
$sin(c)=\frac{1}{2ac}\sqrt{(2ac+a^2-b^2+c^2)(2ac-a^2+b^2-c^2)}$
$sin(c)=\frac{1}{2ac}\sqrt{((a+c)^2-b^2)(b^2-(a-c)^2)}$
(IV) $sin(c)=\frac{1}{2ac}\sqrt{(a+c-b)(a+c+b)(b+a-c)(b-a+c)}$
$a+b+c=2p$, where $p$ is half the perimeter of the triangle. Thus, back to equation IV:
$sin(c)=\frac{1}{2ac}\sqrt{(2p-2b)(2p)(2p-2c)(2p-2a)}$
(V) $sin(c)=\frac{2}{ac}\sqrt{p(p-a)(p-b)(p-c)}$
Plugging V into I:
$h_a=(c)sin(c)$
$h_a=c\frac{2}{ac}\sqrt{p(p-a)(p-b)(p-c)}$
$h_a=\frac{2}{a}\sqrt{p(p-a)(p-b)(p-c)}$
Repeating these steps with the other two heights of triangle ABC,
$h_b=\frac{2}{b}\sqrt{p(p-a)(p-b)(p-c)}$,
$h_c=\frac{2}{c}\sqrt{p(p-a)(p-b)(p-c)}$.
$b_b=\frac{2}{a+c}\sqrt{acp(p-b)}$
$b_c=\frac{2}{a+b}\sqrt{abp(p-c)}$
Where $2p=a+b+c$ ($p$ is half the perimeter of the triangle)
$b_b=\frac{2}{|a-c|}.\sqrt{ac(p-a).(p-c)}$
$b_c=\frac{2}{|a-b|}.\sqrt{ab(p-a).(p-b)}$
Where $2p=a+b+c$ ($p$ is half the perimeter of the triangle)
Find Practice Tests in the following link:
SAT Practice Tests - Metric Relations in Triangles and
Additional Practice Tests - Metric Relations in Triangles
(IV) $sin(c)=\frac{1}{2ac}\sqrt{(a+c-b)(a+c+b)(b+a-c)(b-a+c)}$
$a+b+c=2p$, where $p$ is half the perimeter of the triangle. Thus, back to equation IV:
$sin(c)=\frac{1}{2ac}\sqrt{(2p-2b)(2p)(2p-2c)(2p-2a)}$
(V) $sin(c)=\frac{2}{ac}\sqrt{p(p-a)(p-b)(p-c)}$
Plugging V into I:
$h_a=(c)sin(c)$
$h_a=c\frac{2}{ac}\sqrt{p(p-a)(p-b)(p-c)}$
$h_a=\frac{2}{a}\sqrt{p(p-a)(p-b)(p-c)}$
Repeating these steps with the other two heights of triangle ABC,
$h_b=\frac{2}{b}\sqrt{p(p-a)(p-b)(p-c)}$,
$h_c=\frac{2}{c}\sqrt{p(p-a)(p-b)(p-c)}$.
Stewart's Theorem
In triangle ABC (figure below), a segment (with size "x") connects vertex B to a point on side AC:
By the Law of Cosines:
$c^2=m^2+x^2-2(mx)cos(s)$
Multiplying both sides of this equation by $n$:
$c^2=m^2+x^2-2(mx)cos(s)$
Multiplying both sides of this equation by $n$:
(I) $nc^2=nm^2+nx^2-2(mnx)cos(s)$
Also by the Law of Cosines:
$a^2=n^2+x^2-2(nx)cos(180-s)$
Multiplying both sides of this equation by $m$, and considering that $cos(180-s)=-cos(s)$:
$a^2=n^2+x^2-2(nx)cos(180-s)$
Multiplying both sides of this equation by $m$, and considering that $cos(180-s)=-cos(s)$:
(II) $ma^2=mn^2+mx^2+2(mnx)cos(s)$
Adding equations I and II:
$nc^2+ma^2=nm^2+nx^2+mn^2+mx^2$
$nc^2+ma^2=nm(m+n)+x^2(m+n)$
Since $m+n=b$,
$nc^2+ma^2=nmb+x^2b$ (Stewart's Theorem)
Since $m+n=b$,
$nc^2+ma^2=nmb+x^2b$ (Stewart's Theorem)
$nc^2+ma^2-x^2b=nmb$
$\frac{c^2}{mb}+\frac{a^2}{nb}-\frac{x^2}{nm}=1$ (Stewart's Theorem)
Medians
In the case of a median, $m=n=\frac{b}{2}$ (figure below).
According to Stewart's Theorem
$4x^2=2c^2+2a^2-b^2$
$x=\frac{1}{2}\sqrt{2(a^2+c^2)-b^2}$.
According to Stewart's Theorem
$nc^2+ma^2-x^2b=nmb$
$\frac{b}{2}c^2+\frac{b}{2}a^2-x^2b=\frac{b}{2}\frac{b}{2}b$
$2bc^2+2ba^2-4x^2b=b^3$
$2c^2+2a^2-4x^2=b^2$$\frac{b}{2}c^2+\frac{b}{2}a^2-x^2b=\frac{b}{2}\frac{b}{2}b$
$2bc^2+2ba^2-4x^2b=b^3$
$4x^2=2c^2+2a^2-b^2$
$x=\frac{1}{2}\sqrt{2(a^2+c^2)-b^2}$.
Interior Angle Bisector
Consider triangle ABC (figure below), in which $b_a$ is the bisector of interior angle $\hat{A}$.
Using the Interior Angle Bisector Theorem and the Stewart's Theorem it is possible to prove that the interior bisectors measure:
$b_a=\frac{2}{b+c}\sqrt{bcp(p-a)}$
Using the Interior Angle Bisector Theorem and the Stewart's Theorem it is possible to prove that the interior bisectors measure:
$b_a=\frac{2}{b+c}\sqrt{bcp(p-a)}$
$b_c=\frac{2}{a+b}\sqrt{abp(p-c)}$
Where $2p=a+b+c$ ($p$ is half the perimeter of the triangle)
Exterior Angle Bisector
Consider triangle ABC (figure below), in which $b_a$ is the bisector of exterior angle $\hat{A}$.
Using the Exterior Angle Bisector Theorem and the Stewart's Theorem it is possible to prove that the exterior bisectors measure:
$b_a=\frac{2}{|b-c|}.\sqrt{bc(p-b).(p-c)}$
Using the Exterior Angle Bisector Theorem and the Stewart's Theorem it is possible to prove that the exterior bisectors measure:
$b_a=\frac{2}{|b-c|}.\sqrt{bc(p-b).(p-c)}$
$b_c=\frac{2}{|a-b|}.\sqrt{ab(p-a).(p-b)}$
Where $2p=a+b+c$ ($p$ is half the perimeter of the triangle)
Solved SAT Practice Tests
Find Practice Tests in the following link:
SAT Practice Tests - Metric Relations in Triangles and
Additional Practice Tests - Metric Relations in Triangles
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