Question

In figure, $\mathrm{\angle }B<\mathrm{\angle }A$and $\mathrm{\angle }C<\mathrm{\angle }D$. Show that $AD<BC$.

Answer 1

Hint: Solve the problem by using the theorem of a triangle that is the opposite side of the higher angle is always greater, and the side opposite to the lowest angle is always the shortest.

Complete answer:
Given:
In $\mathrm{\Delta }OAB$, $\mathrm{\angle }B<\mathrm{\angle }A$ and in $\mathrm{\Delta }OCD$, $\mathrm{\angle }C<\mathrm{\angle }D$.
From $\mathrm{\Delta }OAB$ by the theorem of side opposite to larger angle is longer, opposite side of $\mathrm{\angle }B$ is less than opposite side of $\mathrm{\angle }A$.
$\Rightarrow OA<OB$…...(1)
From $\mathrm{\Delta }OCD$, the opposite side of $\mathrm{\angle }C$ is less than the opposite side of $\mathrm{\angle }D$.
$\Rightarrow OD<OC$.…..(2)
Add both equations (1) and (2).
$\begin{array}{c}OA+OD<OB+OC\\ AD<BC\end{array}$
Therefore, side AD is shorter than the side BC.

Note: While applying the theorem always remember to choose the opposite side of the larger angle, not adjacent side. This is a common error that is usually done by students so to eliminate this error remembers the correct concept of the theorem.