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.

hypotenuse leg 1

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 DEFContinue reading…

 

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

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

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.

parabola

 

Suppose that the coordinates of the focus is (0,p) where p > 0, then the directrix is y = -p (can you see why?).

From here we can see that PQ = |y + p| and PF = \sqrt{x^2 + (y - p)^2 }. Since PQ is equal to PF,

|y + p| = \sqrt{x^2 + (y - p)^2 }

Squaring both sides, we have

(y + p)^2 = x^2 + (y-p)^2
y^2 + 2py + p^2 = x^2 + y^2 - 2py + y^2
2py = x^2 - 2py
4py = x^2
y = \frac{1}{4p}x^2.

This is the equation of the parabola with focus (0,p) and directrix y = -p.