Given
∠ABC is an inscribed angle
Find out the m∠BDA and m∠BCA .
To proof
First find the value of the central angle ( intercepted arc measure BA)
∠BOA = 360° - 250°
= 110 °
Thus the intercepted arc AB is of measure 110°
FORMULA
[tex]inscribed\ angle =\frac{1}{2} in tercepted\ arc \ measure[/tex]
thus putting the value in the above equation
we get
[tex]\angle BDA=\frac{1}{2}(110^{\circ})[/tex]
∠BDA = 55°
Now find out ∠BCA
In the quadilateral AOBC
As shown in the diagram AC & BD are tangent
thus
∠CAO = 90°
∠CBO = 90°
As we know the sum of a quadilateral is 360°.
thus
∠AOB + ∠CAO + ∠CBO +∠ BCA = 360°
Put the value as mentioned above
110° +90° + 90° +∠BCA = 360°
∠BCA = 360° - 290°
∠BCA = 70°
Hence proved
