Answer:
[tex]{\angle}BEC=62^{\circ}[/tex]
Step-by-step explanation:
Given:It is given that Chords AB and CD intersect each other at point E.
To find: The measure of ∠BEC.
Construction: Join AC.
Solution:
It is given that Chords AB and CD intersect each other at point E.
Now, we know that the inscribed angle is the half of the intercepted arc, thus
[tex]{\angle}ACD={\frac{1}{2}}(AD)[/tex]
[tex]{\angle}ACD={\frac{1}{2}}(54^{\circ})[/tex]
[tex]{\angle}ACD=27^{\circ}[/tex]
And, [tex]{\angle}CAB={\frac{1}{2}}(CB)[/tex]
[tex]{\angle}CAB={\frac{1}{2}}(70^{\circ})[/tex]
[tex]{\angle}CAB=35^{\circ}[/tex]
Now, in ΔAEC, we have
[tex]{\angle}CAE+{\angle}ACE+{\angle}AEC=180^{\circ}[/tex] (Angle sum property of triangles)
[tex]35+27+{\angle}AEC=180[/tex]
[tex]{\angle}AEC=118^{\circ}[/tex]
Also, using the straight line property, we have
[tex]{\angle}AEC+{\angle}CEB=180^{\circ}[/tex]
[tex]118+{\angle}CEB=180[/tex]
[tex]{\angle}CEB=62^{\circ}[/tex]
Therefore, the measure of [tex]{\angle}BEC[/tex] is [tex]62^{\circ}[/tex].
