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In triangle $ABC,$ $D$ is a point on $\overline{BC}$ such that $BD = DC = DA.$ If $\angle ABC = 45^\circ,$ then how many degrees are in $\angle ACB$?

 Aug 5, 2023
 #1
avatar+753 
0

I think the answer is 45 degrees, because angle abc is 45, and because BD = BC = BA, I think it is an isosceles. I'm not sure. 

 Aug 5, 2023
 #4
avatar+126978 
0

Good job, history !!!

 

cool cool cool

CPhill  Aug 5, 2023
 #2
avatar+126978 
+1

By SAS, triangle BDA  is similar to  triangle CDA

Angle ABD = Angle ACD  = Angle  ACB = 45

 

cool cool cool

 Aug 5, 2023
 #3
avatar+753 
0

Nice Job CPhiII. I was just guessing cuz I had no proof. 

history  Aug 5, 2023
 #5
avatar+753 
0

Also, welcome back.  (Check the message I sent u xD)

history  Aug 5, 2023
 #6
avatar+183 
+3

Another way would be to note that ADB and ADC are congruent, which proves that the angle is 45 as well.

plaintainmountain  Aug 5, 2023
 #7
avatar+126978 
0

Thanks, plaintainmountain....I didn't think of that !!!

 

 

cool cool cool

CPhill  Aug 5, 2023
 #8
avatar+753 
+1

Nice thinking!

history  Aug 5, 2023

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