Verify that in congruent triangles, the bisects of congruent angles are congruent segments

To prove that, in congruent triangles, the bisectors of congruent angles are themselves congruent segments.

Let ABC  and DEF be congruent triangles

And let CC' be the angle bisector of of angle C ....and let FF' be the angle bisector of F

Then, since angle C = angle F, then the bisected angles BCC' and EFF' will also be equal.

And BC = EF.  And angle B = angle E.

Thus, by ASA, triangle BCC' = triangle EFF'.

Thus, the angle bisectors (segments) CC' and FF' are equal

(I think this is something on the order of what you might need...I hope)

The idea is that halves of equal are equal.

The bisector of an angle produces two angles, each one-half the size of the original angle.

Since the original angles are congruent, they are equal in size.

Haves of two equal-sized angles are equal in size.

Therefore, the angles created by the bisecting of two congruent angles are themselves congruent.

Geno3141 can you please explain it like for example

To prove that...

Different teachers want different types of proofs. I will use "m(∠1)" to mean "the measure of angle 1", which is a number, therefore it can be divided by 2 and can be found equal to another number.

Assume that ∠A ≅ ∠X  

     with  ray(AB) bisecting ∠A into the two smaller (inside) angles ∠1 and ∠2

       and ray(XY) bisecting ∠X into the two smaller (inside) angles ∠3 and ∠4.

To show that ∠1 ≅ ∠3.

Since  ∠A ≅ ∠X, m(∠A) = m(∠X).    (If anlges are congruent, they must have the same measure.)

Multiplying both sides by ½:   ½m(∠A) = ½m(∠X).

But, since ray(AB) bisects ∠A into ∠1 and ∠2, m(∠1) = ½m(∠A)

and, since ray(XY) bisects ∠X into ∠3 and ∠4, m(∠3) = ½m(∠X)

Since   ½m(∠A) = ½m(∠X),  m(∠1) = m(∠3).

Since   m(∠1) = m(∠3),  ∠1 ≅ ∠3.

If this doesn't make sense, post back.

CPhill, thank you very much. This is the easiest and right way to do it. This was the only exercise that I wasn't understanding,I have an exam tomorrow. Thank you for your help.

i cant continue to write my proof.Ok triangle ABC is congruent to triangle A'B'C'. Segment AD bisects angle BAC

and segment A'D' bisects angle B'A'C'. Proof segment AD is congruent to A'D'.

 Because triangle ABC is congruent to triangle A'B'C' (given)

so angle BAC = angle B'A'C'  angle BCA = angle B'C'A'$$\Rightarrow$$ angle DCA = angle D'C'A' (corresponding angles of two congruent triangles are equal)

and   segment AC = segment A'C' (corresponding sides of two congruent triangles are equal)

because AD bisects angle BAC (given)

so angle BAD = angle CAD =1/2 angle BAC (angle bisect theorem)

same, angle B'A'D' = angle  C'A'D' =1/2 angle B'A'C'

because angle BAC = angle B'A'C'  (proved)

 1/2 angle BAC =1/2 angle B'A'C  $$\Rightarrow$$ angle DAC =angle D'A'C'

In the triangle DCA and triangle D‘C'A'

angle DAC =angle D'A'C'

segment AC = segment A'C'

angle DCA = angle D'C'A' 

so triangle DCA congruent to  triangle D‘C'A' (ASA) (angle -side-angle )

so segment AD is congruent to segment A'D' (corresponding sides of two congruent triangles are equal)I 

i wish that work. And left side the image have something wrong. i am really sorry. Thank you!