Gideon has prepared the following two-column proof below. He is given that ∠OLN ≅ ∠LNO and he is trying to prove that ΔOLE ≅ ΔONE.
Step Statement Reason 1 ∠OLN ≅ ∠LNO Given 2 Draw OE as a perpendicular bisector to LN by Construction 3 m∠LEO = 90° Definition of a Perpendicular Bisector 4 m∠NEO = 90° Definition of a Perpendicular Bisector 5 LE ≅ EN Definition of a Perpendicular Bisector 6 OL ≅ ON CPCTC 7 ∠LEO ≅ ∠NEO Substitution Property of Equality 8 ΔOLE ≅ ΔONE Angle-Side-Angle (ASA) Postulate I am thinking Angle Side Angle is wrong but I am not sure
I cannot see the image, it requires an account, could you take a screenshot and upload it here?
sure hold on a sec
It is literally the last question on my test and I am stumped
I believe you are correct
I thought so to however I forgot to post something. Gideon made two errors in the proof. Identify and correct the errors.
I think 5 is wrong not sure though
Do you know of anybody who could solve it? If so let them know please.
not anyone on but if they come on @dude @563blackghost
two great mafers right der^
Thanks I will message them.
I will give a medal to whoever answers this. I and desperate as my test was due last week
I know six is wrong because CPCTC is Corresponding Parts of Congruent Triangles are Congruent. I don't think that works here.
Thanks for the help. I submitted the test and I am waiting for the results
Where is point E?
point is in the middle of line NL you cant see it in the picture but point E is made by construction
it is what allows for the line bisector
Mmm the visual would help ;-;
I don't need the help anymore as I tried to figure it out and I submitted it. However I can send you the pic to see what you think the answer is?
Ye I know, but I thought maybeh you would still like an explanation for it >.< pls do send a pic
Alright I will give me a sec
Here it is
I would agree with your answer, the final proven triangle is named as \(\bf{\angle OLE}\) by ASA, but we did not prove it in that way. He never determined that \(\angle O = \angle O\) instead it is proven by SAS Theorem. Considering that LO = NO by CPTCTC, and given that \(\angle L\) = \(\angle N\) with two equal lengths from the bisector. So side-angle-side.
CPCTC does apply. You proved an angle, then a side, and then another angle, so you had proved a triangle already, meaning the other not mentioned sides are equal.
Isn't it the Transitive Property of Equality? What I said in my answer was what you mentioned about the Side Angle Side and that the Substitution property of equality should have been the Transitive.
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