SOLUTION
Consider the image given below,
Since the diagonals of a kite are perpendicular, hence the angles at the center are 90 degrees.
Hence
From triangles of the shorter diagonals, triangle, ADC
[tex]\angle DAE=\angle DCE\text{ (bases angles of isosceles triangle)}[/tex]
Then
[tex]\begin{gathered} \angle ADE+\angle AED+\angle DAE=180^0(\text{ sum of angles in a triangle)} \\ \text{Where } \\ \angle ADE=48^0,\angle AED=90^0,\angle DAE=\angle DCE^{} \end{gathered}[/tex]
substituting the values we have
[tex]\begin{gathered} 48^0+90^0+\angle DCE=180^0 \\ 138^0+\angle DCE=180^0 \end{gathered}[/tex]
Then, subtracting 138 from both sides we have
[tex]\begin{gathered} \angle DCE=180^0-138^0 \\ \angle DCE=42^0 \end{gathered}[/tex]
Hence
The measure of angle DCE= 42 degrees
Similarly, considering the triangle CAB
[tex]\begin{gathered} \angle BCE=\angle BAE(\text{ base angle of an isosceles triangle )} \\ \text{Where } \\ \angle BCE=67^0 \end{gathered}[/tex]
Then
[tex]\angle BCE+\angle BAE+\angle ABC=180^0(\text{ sum fo angles in a triangle CAB)}[/tex]
Hence
[tex]\begin{gathered} 67^0+67^0+\angle ABC=180^0 \\ 134^0+\angle ABC=180^0 \end{gathered}[/tex]
Subtracting 134 from both sides, we have
[tex]\begin{gathered} \angle ABC=180^0-134^0 \\ \angle ABC=46^0 \end{gathered}[/tex]
Hence
m < ABC =46°
Answer: m
