Triangles and parallel lines problem

AI Thread Summary
The discussion revolves around a geometry problem involving triangles and parallel lines, specifically focusing on angle congruences. It is established that angles 1, 2, 3, and 4 are congruent due to vertical angles and given congruences. To prove that line AC is parallel to line BD, it is necessary to demonstrate the congruence of angle 4 with angle 6 or angle 5 with angle 3. The user struggles with expressing the measures of angles 3 and 5 correctly, leading to confusion in their calculations. The conversation highlights the importance of correctly identifying angle relationships to solve the problem.
lingping7
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The picture has the problem and my attempts, I need guidance.
Thanks in advance
 

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Okay, you are given that angle 1 is congruent to angle 4 and angle 2 is congruent to angle 3 (you should say that in the proof). Angle 1 is congruent to angle 2 by "VOA" (vertical angles) so all of angles 1, 2, 3, and 4 are congruent to one another. To prove that AC is parallel to BD, you need to show that angle 4 is congruent to angle 6 or that angle 5 is congruent to angle 3 (alternate interior angles).
 
Yes that is true, I'm able to get angle 5 = to angle 6, but can't equate either of these angles to angle 1,2,3,4
 
How can you express the measure of angle 3? Can you express the measure of angle 5 in the same way?
[edit] I think I was transposing the 4 and the 6. >_<
 
Last edited:
Angle 3 = 180-angle6-angle2
Angle 5 = 180-angle4-angle1
I could say Angle 5 = 180-angle2-angle2
I still don't get it :(
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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