Proofs from The Book - Page 3 of 57 - The most elegant axioms, theorems, and proofs in school mathematics.

Another Proof of the Hypotenuse Leg Theorem

Posted on 10 April, 2016 by Math Proofs Leave a comment

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

Proof that for all real numbers a, |-a| = |a|

Posted on 20 February, 2016 by Math Proofs Leave a comment

We can think of the absolute value of a number as its distance from 0. So, the absolute value of a, which is denoted by $|a|$ is always greater than 0. In this post, we are going to prove that for all real numbers a, |-a| = |a|.

There are two possible cases: $a \geq 0$ and $latex b < 0$. (i) For $latex a \geq 0$, $latex since - a \leq 0$ $latex |-a| = -(-a) = a$ (since a is negative, we negate it to make it positive) $latex |a| = a$ (ii) For $latex a < 0$, since $latex - a > 0$,

$|-a|$ = $-a$ (since a is negative, we negate it to make it positive)
$|a| = -a$

By (i) and (ii), for any real number a,

$|-a| = |a|$ .

Derivation of the Equation of the Parabola

Posted on 20 January, 2016 by Math Proofs Leave a comment

A circle is a locus of points whose distance from a fixed point is a constant. A parabola can also be described as a locus of points whose distance from a fixed point and a fixed line not passing through that point is a constant. An example of a parabola is shown below.

In the figure below, point $F$ is called the focus of the parabola and line $l$ is called its directrix. The vertex of the parabola is at the origin O and the $x$ -axis the perpendicular bisector of $FH$ . If we take any point $P(x,y)$ on the parabola, draw $FP$ , and draw $PQ$ perpendicular to line $l$ where Q is line l, then the distance between $F$ and $P$ and $P$ and $Q$ are equal.