Answer:
[tex]x=34^\circ[/tex]
Step-by-step explanation:
[tex]\text{1. }\angle\text{ABD}=\angle\text{ADB}\text{ and }\angle\text{BCD}=\angle\text{BDC}=x^\circ\ \ \ [\text{Base angles of isosceles triangle}\\\text{}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad\quad\text{are equal.}][/tex]
[tex]\text{2. }\angle\text{ABD}+\angle\text{ADB}+\angle\text{A}=180^\circ\ \ \ [\text{Sum of angles of triangle is 180}^\circ.]\\\text{or, }\angle\text{ABD}+\angle\text{ABD}+44^\circ=180^\circ\\\text{or, }2\angle\text{ABD}=136^\circ\\\text{or, }\angle\text{ABD}=68^\circ[/tex]
[tex]\text{3. }\angle\text{C}+\angle\text{BDC}=\angle\text{ABD}\ \ \ [\text{An exterior angle of a triangle is equal to the}\\\text{}\qquad\qquad\qquad\qquad\qquad\qquad\quad\text{sum of the opposite interior angles.]}\\\text{or, }x^\circ+x^\circ=68^\circ\\\text{or, }2x^\circ=68^\circ\\\text{or, }x^\circ=34^\circ[/tex]
