Question

In the given figure, $AB$ 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

Construction $\to$ Join A & C

Given

AB $\to$ smallest

CD $\to$ longest

$\begin{array}{cc}In\Delta ADC,& \\ CD>AD& \left[CDislongest\right]\\ \angle DAC>\angle DAC\to \left(i\right)& \left[anglesoppositesideisgreater\right]\\ In\Delta ABC,& \\ BC>AB& \left[ABissmallest\right]\\ \angle BAC>\angle ACB\to \left(ii\right)& \left[anglesoppositesideisgreater\right]\\ Adding\left(i\right)\&\left(ii\right),weget& \\ \angle DAC+\angle BAC>\angle DCA+\angle ACB& \\ \angle A>\angle CHenceproved.& \\ \left(ii\right)Toprove\Rightarrow \angle B>\angle D& \\ In\Delta ABD,& \\ AD>AB& \left[ABissmallest\right]\\ \angle ABO>\angle AOB\to \left(i\right)& \left[anglesoppositetosideisgreater\right]\\ In\Delta BDC,& \\ CD>BC& \\ \angle DBC>\angle BDC\to \left(ii\right)& \\ Adding\left(i\right)\&\left(ii\right),weget& \\ \angle ABD+\angle DBC>\angle ADB+\angle BDC& \\ \angle B>\angle Dhenceproved.& \end{array}$