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.

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

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