Post Number:#2 by dshasan » Sat Jan 14, 2017 12:37 am
$Claim 1:$ In a trapezoid $ABCD (AB \parallel CD), AB + CD = 2EF $ where $E,F$ are the midpoints of $AD, BC$
$Proof:$ Extend $AD, BC$ so that they meet at $X$. Now, let $XA = a, XB = b, AE = ED = c, BF = FC = d$.
Now, $\frac{AB}{EF} = \frac{a}{a+c}$ and $\frac{EF}{CD} = \frac{a+c}{a+2c}$. From this two equation, we get, \[ AB + CD = 2EF\]
Now, segment $XY$ is largest when $XY$ passes through the midpoints of $AD$ and $BC$. Let $m,n$ be the midpoints of $AD$ and $BC$ respectively. SO, $Xm = \frac{1}{2} AD$ and $Yn = \frac{1}{2} BC$.
Therefore, $XY = Xm + mn + Yn = \frac{1}{2} ( AB + BC + CD + DA)$. For all other $XY$, it is less than half of the perimeter of $ABCD$
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
- Charles Caleb Colton