# Introduction

I have compiled these formulas from a list of different places and have made sure to include most of the important formulas in geometry. Try to prove most of these theorems by using proofs.

# Angles

These types angles are what you should already know. The Alternate-Interior angles, Alternate-Exterior Angles, Vertical Angles, and Corresponding Angles are always congruent. Treat the parallel line as a line by summing the angles along it to 180 degrees. Same goes for same side angles.

# Power of a Point

1. The two lines are secants of the circle and intersect inside the circle (figure on the left). In this case, we have $$AE\cdot CE = BE\cdot DE$$.
2. One of the lines is tangent to the circle while the other is a secant (middle figure). In this case, we have $$AB^2 = BC\cdot BD$$.
3. Both lines are secants of the circle and intersect outside of it (figure on the right). In this case, we have $$CB\cdot CA = CD\cdot CE.$$

# Triangle Similarity

There are 3 main ways to prove that 2 triangles are similar to one another.

1. AA-Similarity: This is just proving if 2 angles in 2 triangles are congruent, then they are similar, thus, they have the same ratios of their corresponding sides.

2. SSS-Similarity: This can also be used to find similar triangles, though it is quite uncommon as SSS can also imply congruence.

3. SAS-Similarity: 2-sides and 1 angle leads to common ratios of sides.

Kite - $$K=\frac{d_1\cdot d_2}{2}$$ where the $$d$$s represent the lengths of the diagonals of the kite.
Parallelogram - $${K=bh}$$, where $$b$$ is the base and $$h$$ is the height to that base.
Trapezoid - $$K=\frac{b_1+b_2}{2}\cdot h$$, where the $$b$$s are the parallel sides and $$h$$ is the distance between those bases.
Rhombus - a special case of a kite and parallelogram, so either formula may be used here.
Rectangle - $${K=lw}$$, where $$l$$ is the length of the rectangle and $$w$$ is the width. (This is a special case of the formula for a parallelogram where the height and a side happen to coincide.)
Square - $$K=s^2$$, where $$s$$ is the length of a side.
Any quadrilateral - $$K=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left(\dfrac{B+D}{2}\right)}$$, where $$s$$ is the semiperimeter, $$a$$, $$b$$, $$c$$, and $$d$$ are the side lengths, and $$B$$ and $$D$$ are the measures of angles $$B$$ and $$D$$, respectively.
Cyclic quadrilateral - $$K=\sqrt{(s-a)(s-b)(s-c)(s-d)}$$ where $$s$$ is the semiperimeter and $$a$$, $$b$$, $$c$$, and $$d$$ are the side lengths. (This is a special case of the formula for the area of any quadrilateral; $$\cos^2\left(\dfrac{B+D}{2}\right)=0$$.)

# Area of Polygons

The area of any regular polygon can be found as follows:

Inscribe the figure, with $$n$$ sides of length $$s$$, in a circle and draw a line from two adjacent vertices to the circumcenter. This creates a triangle that is $$\frac{1}{n},$$ of the total area (consider the regular octagon below as an example).

Drawing the apothem creates two right triangles, each with an angle of $$\frac{180}{n}^{\circ}$$ at the top vertex. If the polygon has side length $$s$$, the height of the triangle can be found using trigonometry to be of length $$\frac s2 \cot \frac{180}{n}^{\circ}$$.

The area of each triangle is $$\frac12$$ the base times the height, which can also be expressed as $$\frac{s^2}{4} \cot\frac{180}{n}^{\circ}$$ and the area of the entire polygon is $$\frac{n\cdot s^2}{4} \cot\frac{180}{n}^{\circ}$$.

# Area of Triangles and More Rules

There are many ways to find the area of a triangle. In all of these formulae, $${K}$$ will be used to indicate area.

$$K=\frac{bh}{2}$$ where $$b$$ is a base and $$h$$ is the altitude of the triangle to that base. Heron’s formula: $$K=\sqrt{s(s-a)(s-b)(s-c)}$$, where $$a, b$$ and $$c$$ are the lengths of the sides and $$s$$ is the semi-perimeter $$s=\frac{a+b+c}{2}$$. $$K=rs$$, where $$r$$ is the radius of the incircle and s is the semi-perimeter. $$K=\frac{ab\sin{\theta}}{2}$$ where $$a$$ and $$b$$ are adjacent sides of the triangle and $$\theta$$ is the measure of the angle between them. $$K=\frac{abc}{4R}$$, where $$a,b,c$$ are the lengths of the sides of the triangle and $$R$$ is the circumradius. $$\frac{1}{K}=4\sqrt{H(H-h_a^{-1})(H-h_b^{-1})(H-h_c^{-1})}$$, where $$H=\frac{(h_a^{-1}+h_b^{-1}+h_c^{-1})}{2}$$ and the triangle has altitudes $$h_a$$, $$h_b$$, $$h_c$$.

The Triangle Inequality says that in a nondegenerate triangle $$ABC$$: $$AB + BC > AC$$
$$BC + AC > AB$$
$$AC + AB > BC$$

# Angle Bisector Theorem

The Angle Bisector Theorem states that given triangle $$\triangle ABC$$ and angle bisector AD, where D is on side BC, then $$\frac cm = \frac bn$$. Likewise, the converse of this theorem holds as well.

# Uncommon 3-D Areas

Simplify these problems to 2-D problems and sum them up to find Surface Area.

# Uncommon 3-D Formulas

Euler’s Polyhedral Formula Let $$P$$ be any convex polyhedron, and let $$V$$, $$E$$ and $$F$$ denote the number of vertices, edges, and faces, respectively. Then $$V-E+F=2$$.

Menelaus’ Theorem

A necessary and sufficient condition for points $$P, Q, R$$ on the respective sides $$BC, CA, AB$$ (or their extensions) of a triangle $$ABC$$ to be collinear is that $$BP\cdot CQ\cdot AR = PC\cdot QA\cdot RB$$

Ptolemy’s Theorem

Given a cyclic quadrilateral $$ABCD$$ with side lengths $${a},{b},{c},{d}$$ and diagonals $${e},{f}$$:$ac+bd=ef.$
Stewart’s Theorem

Given a triangle $$\triangle ABC$$ with sides of length $$a, b, c$$ opposite vertices $$A$$, $$B$$, $$C$$, respectively. If cevian $$AD$$ is drawn so that $$BD = m$$, $$DC = n$$ and $$AD = d$$, we have that $$b^2m + c^2n = amn + d^2a$$. (This is also often written $$man + dad = bmb + cnc$$, a form which invites mnemonic memorization, e.g. “A man and his dad put a bomb in the sink.”)

Ceva’s Theorem Let $$ABC$$ be a triangle, and let $$D, E, F$$ be points on lines $$BC, CA, AB$$, respectively. Lines $$AD, BE, CF$$ are concurrent if and only if

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

where lengths are directed. This also works for the reciprocal or each of the ratios, as the reciprocal of $$1$$ is $$1$$.

# Trigonometry

Basic Definitions $$\sin A = \frac ac$$ $$\csc A = \frac ca$$ $$\cos A = \frac bc$$ $$\sec A = \frac cb$$ $$\tan A = \frac ab$$ $$\cot A = \frac ba$$ Even-Odd Identities

$$\sin (-\theta) = -\sin (\theta)$$

$$\cos (-\theta) = \cos (\theta)$$

$$\tan (-\theta) = -\tan (\theta)$$

$$\sec (-\theta) = \sec (\theta)$$

$$\csc (-\theta) = -\csc (\theta)$$

$$\cot (-\theta) = -\cot (\theta)$$

Further Conclusions

Based on the above identities, we can also claim that

$$\sin(\cos(-\theta)) = \sin(\cos(\theta))$$

$$\cos(\sin(-\theta)) = \cos(-\sin(\theta)) = \cos(\sin(\theta))$$

This is only true when $$\sin(\theta)$$ is in the domain of $$\cos(\theta)$$. Reciprocal Relations

From the last section, it is easy to see that the following hold:

$$\sin A = \frac 1{\csc A}$$

$$\cos A = \frac 1{\sec A}$$

$$\tan A = \frac 1{\cot A}$$

Another useful identity that isn’t a reciprocal relation is that $$\tan A =\frac{\sin A}{\cos A}$$.

Note that $$\sin^{-1} A \neq \csc A$$; the former refers to the inverse trigonometric functions. Pythagorean Identities

Using the Pythagorean Theorem on our triangle above, we know that $$a^2 + b^2 = c^2$$. If we divide by $$c^2$$ we get $$\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1$$, which is just $$\sin^2 A + \cos^2 A =1$$. Dividing by $$a^2$$ or $$b^2$$ instead produces two other similar identities. The Pythagorean Identities are listed below:

$$\sin^2x + \cos^2x = 1$$ $$1 + \cot^2x = \csc^2x$$ $$\tan^2x + 1 = \sec^2x$$

(Note that the last two are easily derived by dividing the first by $$\sin^2x$$ and $$\cos^2x$$, respectively.) Angle Addition/Subtraction Identities

Once we have formulas for angle addition, angle subtraction is rather easy to derive. For example, we just look at $$\sin(\alpha+(-\beta))$$ and we can derive the sine angle subtraction formula using the sine angle addition formula.

$$\sin(\alpha \pm \beta) = \sin \alpha\cos \beta \pm\sin \beta \cos \alpha$$ $$\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta$$ $$\tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1\mp\tan \alpha \tan \beta}$$

We can prove $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$$ easily by using $$\sin(\alpha + \beta) = \sin \alpha\cos \beta +\sin \beta \cos \alpha$$ and $$\sin(x)=\cos(90-x)$$.

$$\cos (\alpha + \beta)$$

$$= \sin((90 -\alpha) - \beta)$$$$= \sin (90- \alpha) \cos (\beta) - \sin ( \beta) \cos (90- \alpha)$$

$$=\cos \alpha \cos \beta - \sin \beta \sin \alpha$$ Double Angle Identities

Double angle identities are easily derived from the angle addition formulas by just letting $$\alpha = \beta$$. Doing so yields:

\begin{aligned} \sin 2\alpha &=& 2\sin \alpha \cos \alpha\\ \cos 2\alpha &=& \cos^2 \alpha - \sin^2 \alpha\\ &=& 2\cos^2 \alpha - 1\\ &=& 1-2\sin^2 \alpha\\ \tan 2\alpha &=& \frac{2\tan \alpha}{1-\tan^2\alpha} \end{aligned}

Further Conclusions

We can see from the above that

$$\csc(2a) = \frac{\csc(a)\sec(a)}{2}$$ $$\sec(2a) = \frac{1}{2\cos^2(a) - 1} = \frac{1}{\cos^2(a) - \sin^2(a)} = \frac{1}{1 - 2\sin^2(a)}$$ $$\cot(2a) = \frac{1 - \tan^2(a)}{2\tan(a)}$$

Half Angle Identities

Using the double angle identities, we can now derive half angle identities. The double angle formula for cosine tells us $$\cos 2\alpha = 2\cos^2 \alpha - 1$$. Solving for $$\cos \alpha$$ we get $$\cos \alpha =\pm \sqrt{\frac{1 + \cos 2\alpha}2}$$ where we look at the quadrant of $$\alpha$$ to decide if it’s positive or negative. Likewise, we can use the fact that $$\cos 2\alpha = 1 - 2\sin^2 \alpha$$ to find a half angle identity for sine. Then, to find a half angle identity for tangent, we just use the fact that $$\tan \frac x2 =\frac{\sin \frac x2}{\cos \frac x2}$$ and plug in the half angle identities for sine and cosine.

To summarize:

$$\sin \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}2}$$ $$\cos \frac{\theta}2 = \pm \sqrt{\frac{1+\cos \theta}2}$$ $$\tan \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$$

Prosthaphaeresis Identities

(Otherwise known as sum-to-product identities)

$$\sin \theta \pm \sin \gamma = 2 \sin \frac{\theta\pm \gamma}2 \cos \frac{\theta\mp \gamma}2$$ $$\cos \theta + \cos \gamma = 2 \cos \frac{\theta+\gamma}2 \cos \frac{\theta-\gamma}2$$ $$\cos \theta - \cos \gamma = -2 \sin \frac{\theta+\gamma}2 \sin \frac{\theta-\gamma}2$$

Law of Sines

Main article: Law of Sines

The extended Law of Sines states

$$\frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C} = 2R.$$

Law of Cosines

Main article: Law of Cosines

The Law of Cosines states

$$a^2 = b^2 + c^2 - 2bc\cos A.$$

Law of Tangents

Main article: Law of Tangents

The Law of Tangents states that if $$A$$ and $$B$$ are angles in a triangle opposite sides $$a$$ and $$b$$ respectively, then

$$\frac{\tan{\left(\frac{A-B}{2}\right)}}{\tan{\left(\frac{A+B}{2}\right)}}=\frac{a-b}{a+b} .$$ Other Identities

$$e^{i\theta} = \cos \theta + i\sin \theta$$ (This is also written as $$\text{cis }\theta$$) $$|1-e^{i\theta}|=2\sin\frac{\theta}{2}$$ $$\left(\tan\theta + \sec\theta\right)^2 = \frac{1 + \sin\theta}{1 - \sin\theta}$$ $$\sin(\theta) = \cos(\theta)\tan(\theta)$$ $$\cos(\theta) = \frac{\sin(\theta)}{\tan(\theta)}$$ $$\sec(\theta) = \frac{\tan(\theta)}{\sin(\theta)}$$ $$\sin^2(\theta) + \cos^2(\theta) + \tan^2(\theta) = \sec^2(\theta)$$ $$\sin^2(\theta) + \cos^2(\theta) + \cot^2(\theta) = \csc^2(\theta)$$

The two identities right above here were based on identites others posted on this site with a substitution.

$$\cos(2\theta) = (\cos(\theta) + \sin(\theta))(\cos(\theta) - \sin(\theta))$$