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.
+781 Unfortunately I'm stuck on this too. I tried asking for help but I was accused of cheating by another "Guest".
+1005 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!
+781 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.
+1005 Ah, EBA and ACE?
Are you sure about that?
EBA appears to be Scalene... and obtuse. ACE is (assumingly) isoceles and acute.
