Question

Look at the figure. Show that length $AH>$ sum of lengths of $AB+BC+CD$

Answer 1

Step: Comparing the lengths

Observing the figure we see,
$AB+BC+CD=AD$
$[\because AB,BC$ and $CD$ are the parts of $AD]$



Also from figure, $AD$ is part of $AH$.
According to Euclid’s axiom, the whole is greater than the part.
$\Rightarrow AH>AD$
$\therefore AH>AB+BC+CD$
Hence, Proved.