# List of problems in high school geometry

## Problem 1

Let $B_1C_1\parallel BC$
prove that $\overline{BM}=\overline{MC}$. ## Proof:

Clearly \begin{eqnarray} \triangle BKM\sim \triangle C_1KH \tag{1}\\ \triangle CKM \sim \triangle B_1KH \tag{2}\\ \triangle BKC\sim\triangle C_1KB_1 \tag{3} \label{prob1Sim} \end{eqnarray} Let $\overline{B_1H}=x$ and $\overline{HC_1}=y$. Similarity (3) implies that there existence a real $p$ such that \begin{equation} \overline{BK}=p\cdot \overline{KC_1} \tag{*} \end{equation} From (*) and similarity (1) we get \begin{equation} \overline{BM}=p\cdot\overline{HC_1}=py \tag{a} \end{equation} From similarity (3) we get that \begin{equation} \overline{KC}=p\cdot\overline{B_1K} \tag{**} \end{equation} Using similarity (2) and (**) we get \begin{equation} \overline{MC}=p\cdot \overline{B_1H}=px \tag{b} \end{equation} Since $\triangle ABM\sim \triangle AB_1H$ and $\triangle AMC\sim \triangle AHC_1$, we can use equalities (a) and (b) and get $\frac{px}{y}=\frac{MC}{HC_1}=\frac{AM}{AH}=\frac{BM}{B_1H}=\frac{py}{x}$ therefor $px^2=py^2$ and from this we get $x=y$. $\blacksquare$

## Problem 2

Let $AB$ and $AC$ be two chords in a circle with center $O$.
Given: $\overline{AO}$ bisect $\angle BAC$
Prove: $\overline{AB}=\overline{AC}$. ## Proof:

Let $\angle BAO=\alpha$, $\angle OAC=\alpha_0$, $\angle AOB=\beta$, $\angle COA=\beta_0$, $\angle OBA=\gamma$ and $\angle ACO=\gamma_0$. Since $AO$, $OC$ and $OB$ are radii of the same circle, we get $\triangle AOC$ and $\triangle AOB$ are isosceles triangles. Hence, $\alpha_0=\gamma_0$ and $\alpha=\gamma$. The sum of angles in triangle is $180^\circ$, thus $\alpha+\gamma+\beta=180^\circ = \alpha_0+\beta_0+\gamma_0$ By substituting $\gamma_0$ with $\alpha_0$ we get $2\alpha+\beta=2\alpha_0+\beta_0.$ Given (in the question) that $AO$ is a bisector of $\angle BAC$ we get $\alpha =\alpha_0$, hence $\beta=\beta_0$. Now, $\beta=\beta_0$ and $\alpha=\alpha_0$ and $\overline{AO}= \overline{AO}$ (every segment is equal in length to itself), so $\triangle ABO\cong \triangle ACO$ (A.S.A). From the congruency property we immediately get $AB=AC$. $\blacksquare$

# Problem 3

Chord $AB$ in a circle with center $O$ bisects angle $\angle CBA$
Prove: $CB\parallel AO$ ## Proof:

Since $AO$ and $BO$ are radii of the same circle, $\overline{AO}=\overline{BO}, hence \angle OAB = \angle ABO.$ Given that $AB$ bisects $\angle CBO$, we get $\angle ABO = \angle CBA$ The last two equality implies that $\angle OAB = \angle CBA$, so $AO$ and $CB$ has a pair of equal alternate angles, therefore $AO\parallel CB$. $\blacksquare$

## Problem 4

The diameter is the largest chord in a circle.
Given a diameter $AB$ and a non diameter chord $DC$.
Prove: $\overline{DC}<\overline{AB}$. ## Proof:

Since $\overline{OC}=\overline{OB}=\overline{OA}=\overline{OD}$ (all are radii of the same circle), we get $\overline{OC}+\overline{OD}=\overline{AO}+\overline{OB}=\overline{AB}.$ Since the sum of two sides (in length) is always geater than the length of the remaining (third side), we get $\overline{OC}+\overline{OD}>\overline{DC}.$ Therefore $\overline{DC}<\overline{OD}+\overline{OD}=\overline{AB}$. $\blacksquare$

## Problem 5

Let $AB$ and $AC$ be two chords in a circle with center $O$, where $C$ lies on the arc $AB$.
Prove that: $\angle OBA < \angle OCA$. ## Proof:

"Clearly" $\angle CAO > \angle BAO.$ In addition $\overline{OA}=\overline{OB}=\overline{OC}$ (radii of the same circle), hence triangles $\triangle AOB$ and $\triangle AOC$ are isosceles triangles and we get $\angle CAO=\angle OCA$ and $\angle BAO=\angle OBA$. Hence, $\angle OCA = \angle CAO > \angle BAO = \angle OBA$ so $\angle OCA>\angle OBA$. $\blacksquare$