Question

$AB$ and $CD$ are respectively the smallest and longest sides of a quadrilateral $ABCD$.

Show that $\angle A>\angle C$ and $\angle B>\angle D$.

Answer 1

Step $1:$ Proving $\angle A>\angle C$.

Join $AC$ and mark angles as $\angle 1,\angle 2,\angle 3,\angle 4$, we get,

In, $∆ABC$, $AB<BC$ ($AB$ is the smallest side)

$\therefore \angle 4<\angle 2$ (Angles opposite to smaller side is smaller)…….$\left(1\right)$

Again, In, $∆ADC$, $AD<CD$ ($CD$ is the longest side)

$\therefore \angle 3<\angle 1$ (Angles opposite to smaller side is smaller)…….$\left(2\right)$

Adding equations $\left(1\right)$ and $\left(2\right)$, we get,

$\angle 4+\angle 3<\angle 2+\angle 1$ or

$\begin{array}{rcl}\angle 1+\angle 2& >& \angle 3+\angle 4\\ \angle A& >& \angle C(\angle A=\angle 1+\angle 2,\angle C=\angle 3+\angle 4)\end{array}$

Hence, $\angle A>\angle C$ proved.

Step $2:$ Proving $\angle B>\angle D$.

Join $BD$ and mark angles as $\angle 5,\angle 6,\angle 7,\angle 8$, we get,

In, $∆ABD$, $AB<AD$ ($AB$ is the smallest side)

$\therefore \angle 6<\angle 5$ (Angles opposite to smaller side is smaller)…….$\left(3\right)$

Again, In, $∆CBD$, $BC<CD$ ($CD$ is the longest side)

$\therefore \angle 8<\angle 7$ (Angles opposite to smaller side is smaller)…….$\left(4\right)$

Adding equations $\left(3\right)$ and $\left(4\right)$, we get,

$\angle 6+\angle 8<\angle 5+\angle 7$ or

$\begin{array}{rcl}\angle 5+\angle 7& >& \angle 6+\angle 8\\ \angle B& >& \angle D(\angle B=\angle 5+\angle 7,\angle D=\angle 6+\angle 8)\end{array}$

Hence, $\angle B>\angle D$ proved.