Answer:
[tex]m\angle ABE=52^\circ[/tex]
Step-by-step explanation:
Angles and Lines
The figure shows a kite ABCD where the following data is given:
[tex]m\BAE=28^\circ[/tex]
[tex]m\BCE=58^\circ[/tex]
We also know
[tex]BC\cong CD[/tex]
[tex]AB\cong AD[/tex]
Since triangle BCD is isosceles, it follows that:
[tex]m\angle CBE=m\angle CDE[/tex]
And triangles BCD and DCE are congruent.
Thus:
[tex]m\angle BCE=m\angle DCE=58^\circ[/tex]
And also:
[tex]m\angle BCE=58^\circ+58^\circ=116^\circ[/tex]
Since the sum of internal angles of BCD is 180°
[tex]m\angle CDE=m\angle CBE= ( 180 - 116 ) / 2=32^\circ[/tex]
The sum of angles of triangle CDE is 180°, thus
[tex]m\angle CED=180^\circ - 32^\circ - 58^\circ = 90^\circ[/tex]
Angles CED and BEA are vertical (opposite), thus:
[tex]m\angle BCE=m\angle CED= 90^\circ[/tex]
Finally, since the sum of the angles of triangle ABE is 180°
[tex]m\angle ABE= 180 - 90 - m\angle BAE[/tex]
[tex]m\angle ABE= 180^\circ - 90^\circ - 28^\circ = 52^\circ[/tex]
[tex]\mathbf{m\angle ABE=52^\circ}[/tex]
