in triangle ABC the angle bisector of BAC meets BC at D such that AD = AB. Line segment AD is extended to E such that CD equals CE and DBE = BAD show how triangle ACE is isosceles. Also please explain. Thank you.

https://latex.artofproblemsolving.com/d/3/5/d35b00e81a091116526d86f896d59616cfaa2463.png

Link to the image.

Guest Jun 21, 2020

#1**0 **

Unfortunately I'm stuck on this too. I tried asking for help but I was accused of cheating by another "Guest".

gwenspooner85 Jun 21, 2020

#2**-2 **

Hi!

Gwen, and guest, here are a few hints.

From my response:

We notice that triangles ABD and CDE are isosceles. If triangle ACE is indeed isosceles, then . It also appears that triangles ADC and BDE are similar. This is because BDE and ADC are equal, ADB and BAC are also equal. Furthermore,DBE = BAD , and since AE is the angle bisector of BAC, DBE = DAC. Since two angles are congruent, these triangles are similar. Because triangles BDE and ADC are similar, triangles ABD and CDE are too.

This is MY OWN RESPONSE

So I IMPLORE you to not copy this.

Please just try to follow my logic(this is correct, 7/7 and 0.8/1, note this is part of it)

At the very least paraphrase.

Feel free to ask about anything!

hugomimihu Jun 22, 2020

#3**0 **

Thanks for some guidance. I do have some of this problem done but I'm just stuck right now. Hopefully this will help. Also, then wouldn't triangle ACE be similar to EBA?

Also, I'm seeing a lot more relationships between the angles now that you gave me a hint.

gwenspooner85
Jun 22, 2020

#4**-1 **

Ah, EBA and ACE?

Are you sure about that?

EBA appears to be Scalene... and obtuse. ACE is (assumingly) isoceles and acute.

hugomimihu
Jun 22, 2020