Question

Can someone help me with this question

Answer 1

Step-by-step explanation:

In the given circle

$BC \cong BD$ (Radii of same circle)

$\angle BDC \cong \angle BCD.... (1)$

(angles opposite to congruent sides are congruent)

$\angle ABC$ is exterior angle of $\triangle BDC$

$m\angle ABC= m\angle BDC +m\angle BCD\\\therefore 124°= m\angle BDC+m\angle BDC\\.. [from\:(1)] \\\therefore 124°= 2m\angle BDC\\\\\therefore m\angle BDC= \frac{124°}{2}\\\\\therefore m\angle BDC= 62°\\\huge \pink {\boxed {\therefore m\angle ADC= 62°}} \\.. (\because A-B-D)$