Synthetic Geometry

The inscribed angle subtended by the diameter of circle is right angle.

• Inscribed angle is twice the central angle.
• Euclid's Elements, Proposition 20 on Book 3.

• For a cyclic quadrilateral \(\square ABCD\). \[ \overline{AB}\,\overline{CD} + \overline{BC}\,\overline{AD} = \overline{AC}\,\overline{BD}. \]

• For a triangle \(\triangle ABC\) and a middle point \(D\) of the side \(\overline{BC}\): \[ 2\left(\overline{AB}^2 + \overline{AC}^2\right) = \overline{BC}^2 + 4\overline{AD}^2. \]

\begin{align*} \frac{1}{2}\overline{AB}^2 + \frac{1}{2}\overline{AC}^2 &= \frac{1}{4}\overline{BC}^2 + \overline{AD}^2 \\ \frac{\overline{BC}}{2}\overline{AB}^2 + \frac{\overline{BC}}{2}\overline{AC}^2 &= \overline{BC}\frac{\overline{BC}^2}{4} + \overline{BC}\cdot\overline{AD}^2. \end{align*}

• excalidraw:stweart.excalidraw
• If a cevian divides the side with length \(a\) into \(n\) and \(m\): \[ b^2m+c^2n = a(d^2 + mn). \]

• \[ man + dad = bmb + cnc \]
• A man and his dad put a bomb in the sink.

\[ A = \sqrt{s(s-a)(s-b)(s-c)}, \] where \( s \) is the semi-perimeter of a triangle.

• The area can be obtained in two ways:
  • \[ A = rs \]
    • where \(r\) is the inradius, and \(s\) is the semi-perimeter.
  • \[ A = r(s-a) + r(s-b) + r(s-c), \]
    • Using the triple cotangent identity: \[ A = r^2\left(\frac{s-a}{r}\cdot\frac{s-b}{r}\cdot\frac{s-c}{r}\right). \]
• Multiplying two equations yields: \[ A^2 = s(s-a)(s-b)(s-c). \]

• Given a parallelogram \(\square ABCD\): \[ 2AB^2 + 2BC^2 = AC^2 + BD^2. \]

• \( \overline{BM}\perp\overline{AC},\overline{EF}\perp\overline{BC} \implies |\overline{AF}|=|\overline{FD}|\)

• Area \(K\) of a cyclic quadrilateral with side lengths \(a, b, c, d\): \[ K = \sqrt{(s-a)(s-b)(s-c)(s-d)} \]
• where \(s\) is the semiperimeter.

• \[ K = \sqrt{(s-a)(s-b)(s-c)(s-d) - abcd\cos^2\left(\frac{\alpha+\gamma}{2}\right)} \]

The three center of the equilateral triangles subtended from three sides of any triangle forms an equilateral triangle.

Line segment that joins a vertex and a point on the opposite side in a triangle.

• Given three cevians that intersects at one point as follows, not considering the degenerate cases:

\[ \frac{\overline{AF}}{\overline{FB}}\cdot \frac{\overline{BD}}{\overline{DC}}\cdot \frac{\overline{CE}}{\overline{EA}} = 1, \] using the signed length.

• It is a theorem of affine geometry, that is true for any affine plane over any field.

• \[ \frac{\overline{AF}}{\overline{FB}}\cdot\frac{\overline{BD}}{\overline{DC}}\cdot\frac{\overline{CE}}{\overline{EA}} = -1. \]
• where the lengths are signed.
• The lengths can be interpreted as the homothety .

• Given a pair of sets of collinear points, \( (\{A, B, C\}, \{a, b, c\}) \), the intersection points \( X, Y, Z \) are collinear on the Pappus line.

• The three intersections of the three pairs of line segments that connects opposite points of any six points on a conic section is collinear.
• The line that the intersections passes is called the Pascal line of the hexagon.

• Midpoints of parallel lines are collinear to the center.
• How to find the center, major axis and minor axis of an ellipse - Quora

• If the three intersection points of the three pairs of lines through the opposite sides of a hexagon are collinear, then the vertices lie on a conic.

• Any five points uniquely determines a conic.

• The two triangles subtended by the two chords passing through the midpoint (more generally, two points that are equidistant from the midpoint) of any chord, intersect with the original chord at the same distance from the midpoint.
• It is also applicable to any conic section.

The sum of the distance from any point in a equilateral triangle to each edge is equal to the height of the triangle.

A 2-dimensional plot that can represent three variables with constant sum.

• It make use of barycentric coordinate system, thus making it convenient to represent a composition.

• After extending the sides at each vertex by the length of the opposite side, then the six points at the end of lines lie on a circle called Conway circle which is concentric to the incenter.

• The radius of the Conway circle is \(\sqrt{r^2+s^2}\), where \(r\) is the inradius and \(s\) is the semiperimeter.
• This theorem is the special case of the windscreen wiper theorem, which states that by swiveling a point, one can obtain a circle.

\( X_{1} \), \( I \)

\( X_2 \), \( G \)

\( X_3 \), \( O \)

\( X_4 \), \( H \)

\( X_5 \), \( N \)

Center of the circle passing through

• the midpoint of each side
• the foot of each altitude
• and the midpoint between the orthocenter and each vertex.

The circle goes by various names: nine-point circle, Feuerbach's circle, Euler's circle, Terquem's circle

\( X_6 \), \( K \)

Intersection of the symmedians.

\( X_7 \), \( G_e \)

Intersection of the lines connecting each vertex to the point where the incircle touches the opposite side.

\( X_8 \), \( N_a \)

Intersection of the lines connecting each vertex to the point where the excircle touches the opposite side.

\( X_9 \), \( M \)

• Invariant point under affine transformation?

\( X_{10} \), \( S_p \)

Incenter of the medial triangle.

\( X_{11} \), \( F \)

The tangent point of the incircle and the nine-point circle

\( X_{13} \), \( X \)

Locus of point \( P \) in the plane of the reference triangle \( \triangle ABC \) such that, if the reflections of \( P \) in the sidelines of triangle are \( P_a, P_b, P_c \), then the lines \( AP_a, BP_b, CP_c \) are concurrent.

The content of an \(n\)-simplex formed by \(n+1\) points is given as: \[ v_n^2 = \frac{1}{(n!)^22^n}\begin{vmatrix} 2d_{01}^2 & d_{01}^2+d_{02}^2-d_{12}^2 & \cdots & d_{01}^2+d_{0n}^2-d_{1n}^2 \\ d_{02}^2+d_{01}^2-d_{21}^2 & 2d_{02}^2 &\cdots & d_{02}^2+d_{0n}^2-d_{2n}^2 \\ \vdots &\vdots &\ddots &\vdots \\ d_{0n}^2+d_{01}^2-d_{n1}^2 & d_{0n}^2+d_{02}^2-d_{n2}^2 & \cdots & 2d_{0n}^2 \end{vmatrix} \] where \(d_{ij}\) is the distance between \(i\)-th and \(j\)-th points.

Equivalently: \[ v_n^2 = \frac{(-1)^{n+1}}{(n!)^22^n} \begin{vmatrix} 0 & d_{01}^2 & d_{02}^2 & \cdots & d_{0n}^2 & 1 \\ d_{10}^2 & 0 & d_{12}^2 & \cdots & d_{1n}^2 & 1 \\ d_{20}^2 & d_{21}^2 & 0 & \cdots & d_{2n}^2 & 1 \\ \vdots &\vdots & \vdots &\ddots &\vdots & \vdots \\ d_{n0}^2 & d_{n1}^2 & d_{n2}^2 & \cdots & 0 & 1\\ 1 & 1& 1& \cdots & 1 & 0 \end{vmatrix}. \]