In the figure to the right, the isosceles ΔABC with a base AC and measure of ∠B equal to 36° has angle bisector AD drawn through it. Prove that triangles ∆CDA and ∆ADB are isosceles

Refer to the attached image.Given : Triangle ABC is isosceles, measure of ∠B equal to 36° and AD is an angle bisector.To prove: Triangle CDA and ADB are isoscelesProof:Since triangle ABC is isosceles,therefore AB=BCNow, [tex] \angle A=\angle C [/tex](Angles opposite to the equal opposite sides are always equal)Let [tex] \angle A=\angle C = x [/tex]Therefore, by angle sum property which states"The sum of all the angles of a triangle is 180 degrees"[tex] \angle A+\angle B+\angle C=180^{\circ} [/tex][tex] x+36^{\circ}+x=180^{\circ} [/tex][tex] 2x+36^{\circ}=180^{\circ} [/tex][tex] 2x=180^{\circ}-36^{\circ} [/tex][tex] 2x= 72^{\circ} [/tex][tex] x=36^{\circ} [/tex]Hence, [tex] \angle A=\angle C=72^{\circ} [/tex]Since, AD is an angle bisector.Therefore, it divides angle A into two equal parts.Therefore, [tex] \angle BAD=\angle DAC=36^{\circ} [/tex]Now, consider triangle ABD,here since [tex] \angle ABD=\angle BAD=36^{\circ} [/tex]Therefore, AD = BD"By the converse of the base angles theorem, which states that if two angles of a triangle are congruent, then sides opposite those angles are congruent."Therefore, Triangle ABD is isosceles triangle.Similarly consider triangle ACD,By angle sum property,[tex] \angle ADC+\angle DCA+\angle CAD=180^{\circ} [/tex][tex] 36^{\circ} +72^{\circ} + \angle ADC = 180^{\circ} [/tex][tex] \angle ADC = 72^{\circ} [/tex]Therefore, [tex] \angle ADC=\angle ACD=72^{\circ} [/tex]Therefore, AC = CD"By the converse of the base angles theorem, which states that if two angles of a triangle are congruent, then sides opposite those angles are congruent."Therefore, Triangle ADC is isosceles triangle.