# Another Proof of the Hypotenuse Leg Theorem

Given two triangles, if their hypotenuse are congruent, and one pair of their legs are congruent then the two triangles are congruent. In this post, we are going to prove this theorem.

In the figure below, ABC and DEF are right triangles with right angles at C and F, respectively.

It is given that $latex \overline{AB} \cong \overline{DE}$ and $latex \overline{AC} \cong \overline{DF}$. We are going to prove that $latex \triangle ABC \cong \triangle DEF$.  Continue reading

# Proof that There Are Only Three Regular Tessellations

Tessellation is the process of covering the plane with shapes without gaps and overlaps. Tiles on walls and floors are the most common tessellations.

In this post, we are going to prove that the there are only three regular polygons that can tile the plane. Tiling the plane with regular polygons is called regular tessellation. These are equilateral triangles, squares, and regular hexagons. This proof only requires simple algebra. We will use the notation $latex \{a,b \}$, where $latex a$ is the number angles of the polygon and $latex b$ is the number of vertices that meets at a point. For example, a square has four angles, and at every point on the tessellation, four vertices meet at a point. So, we can represent square as $latex \{ 4,4 \}$. Another example is the regular hexagon. A hexagon has 6 sides, and at every point 3 vertices meet (see figure below), so a hexagon can be represented as $latex \{ 6,3 \}$.  Continue reading

# Sample Proof on Triangle Congruence Part 3

Recall that for real numbers a, b, and c are real numbers, if $latex a = b$ and $latex b = c$, then $latex a = c$. This is called Transitive Property of Equality. This is also the same with congruence. If $latex A$, $latex B$, and $latex C$ are polygons, and if $latex A$ is congruent to $latex B$, and $latex B$ is congruent $latex C$, then $latex A$ is congruent to $latex C$. This is called the Transitive Property of Congruence. We will use this to prove the following problem.

Given: $latex \overline{BE} \cong \overline{DC}$ and $latex \overline{BD} \cong \overline{CA}$.

Prove: $latex \triangle DBE \cong \triangle CAB$.

Proof

It is given that $latex \overline{BE} \cong \overline{DC}$.

Now, by reflexive property, that is a segment is congruent to itself, $latex \overline{BD} \cong \overline{BD}$.  Continue reading