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Metea Valley High School Geometry Formulas

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.

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.

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 \(CB\cdot CA = CD\cdot CE.\)

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; \(\cos^2\left(\dfrac{B+D}{2}\right)=0\).)

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 \(\frac{n\cdot s^2}{4} \cot\frac{180}{n}^{\circ}\).

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 \(ABC\): \(AB + BC > AC\)

\(BC + AC > AB\)

\(AC + AB > BC\)

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.

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

**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 \(1\).

**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.

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

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