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 isosceles

Proof:

Since triangle ABC is isosceles,

therefore AB=BC

Now,

(Angles opposite to the equal opposite sides are always equal)

Let

Therefore, by angle sum property which states

"The sum of all the angles of a triangle is 180 degrees"

Hence,

Since, AD is an angle bisector.

Therefore, it divides angle A into two equal parts.

Therefore,

Now, consider triangle ABD,

here since

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,

Therefore,

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.

1.3

step-by-step explanation:

3x2=6   9x+6

bring down the 6 on both sides

6+6=12

12= 9x

divide by 9

12/9 =

1.3 =x

∡ eia and ∡ kij are congruent

step-by-step explanation: