# Plane Geometry Techniques

#### ✔ Given that orthocenter $H$, incenter $I$, and circumcenter $O$ for a triangle $ABC$, the following statements hold.

(a) If $ABC$ is an acute triangle, $\angle BHC = \pi - \alpha$.

(b) $\angle BIC = \cfrac{\pi + \alpha}{2}$

#### ✔ For a triangle $ABC$, the following holds.

\begin{aligned} \begin{cases} \; a \sin \theta_1 : b \sin \theta_2 = c : d \quad (\theta_1 \ne \theta_2) \\\\ \; a : b = c : d \quad (\theta_1 = \theta_2) \end{cases} \end{aligned}

#### ✔ When $ABCD$ is a convex quadrilateral, concave quadrilateral, or not even a quadrilateral, the following holds.

\begin{aligned} \overline{AC} \perp \overline{BD} \iff \overline{AB}^2 + \overline{CD}^2 = \overline{AD}^2 + \overline{BC}^2 \end{aligned}

#### ✔ (Stewart’s theorem) For a triangle $ABC$, the following holds.

\begin{aligned} a (d^2 + mn) = b^2 m + c^2 n \end{aligned}

#### ✔ Given $D \in \overline{BC}$ and $X \in \overline{AD}$ for a triangle $ABC$, the following holds.

\begin{aligned} \cfrac{\text{Area} (BCX)}{\text{Area} (ABC)} = \cfrac{\overline{DX}}{\overline{DA}} \end{aligned}

#### ✔ (Van Aubel’s theorem) Given $D \in \overline{BC}$, $E \in \overline{AC}$ and $F \in \overline{AB}$ for a triangle $ABC$, if $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ pass through a same point $P$, the following holds.

\begin{aligned} \cfrac{\overline{AP}}{\overline{PD}} = \cfrac{\overline{AE}}{\overline{EC}} + \cfrac{\overline{AF}}{\overline{FB}} \end{aligned}

#### ✔ (Erdos-Mordell) For a triangle $ABC$ and $P$ inside $ABC$, let $PX$, $PY$, and $PZ$ be the perpendiculars from $P$ to the sides of $ABC$. Then, the following holds.

\begin{aligned} \overline{PA} + \overline{PB} + \overline{PC} \geq 2 (\overline{PX} + \overline{PY} + \overline{PZ}) \end{aligned}

#### ✔ For a triangle $ABC$ and $P$ inside $ABC$, at least one of $\angle PAB$, $\angle PBC$, and $\angle PCA$ is $\leq \pi / 6$.

Let $PX$, $PY$, and $PZ$ be the perpendiculars from $P$ to the sides of $ABC$. Then, $\overline{PA} + \overline{PB} + \overline{PC} \geq 2 (\overline{PX} + \overline{PY} + \overline{PZ})$. It implies that one of the followings holds. \begin{aligned} \overline{PB} \geq 2 \overline{PX}, \quad \overline{PC} \geq 2 \overline{PY}, \quad \overline{PA} \geq 2 \overline{PZ} \end{aligned}

Without loss of generality, assume $\overline{PB} \geq 2 \overline{PX}$. Then, \begin{aligned} \sin \angle PBC = \frac{\overline{PX}}{\overline{PB}} \leq \frac{1}{2} \implies \sin \angle PBC \leq \frac{\pi}{6} \end{aligned}

#### ✔ For a convex quadrilateral $ABCD$, find $X$ such that minimize $\overline{XA} + \overline{XB} + \overline{XC} + \overline{XD}$.

By triangle inequality, $\overline{XA} + \overline{XC} \geq \overline{AC}$ and $\overline{XB} + \overline{XD} \geq \overline{BD}$ for the triangles $ACX$ and $BDX$. Therefore, \begin{aligned} \overline{XA} + \overline{XB} + \overline{XC} + \overline{XD} \geq \overline{AC} + \overline{BD} \end{aligned}

Since the equal sign is valid only when $\overline{XA} + \overline{XC} = \overline{AC}$ and $\overline{XB} + \overline{XD} = \overline{BD}$, $X$ is the intersection of $\overline{AC}$ and $\overline{BD}$.

## Reference

[1] Titu Andreescu, 106 Geometry Problems.