Answer:
[tex]m\widehat{BD}=70^{\circ}[/tex]
Step-by-step explanation:
APB is a straight line.
Using liner pair(sum of angles on straight line is 180 degree)
[tex]\angle APF+\angle FPB=180^{\circ}[/tex]
But [tex]\angle APF=110^{\circ}[/tex]
[tex]\therefore 110^{\circ|+\angle FPB=180^{\circ}[/tex]
[tex]\angle FPB=70^{\circ}[/tex]
In triangle FPD, FP=DP (radius of same circle.
So, triangle FPD would be isosceles triangle. FD is base of triangle.
PE is perpendicular to FD.
As we know angle bisector of isosceles triangle make perpendicular to base.
So, line PE bisect angle FPD.
[tex]\therefore \angle FPE=\angle DPE[/tex]
As we found above [tex]\angle FPE=70^{\circ}[/tex]
So, [tex]\angle DPE=70^{\circ}[/tex]
[tex]m\widehat{BD}=70^{\circ}[/tex]
