MHB Can Thales' Theorem Help Me Solve My Trapezoid Equation?

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The discussion revolves around calculating angles E and F in a trapezoid where angles A, B, C, and D are already known. The trapezoid is identified as cyclic and isosceles, leading to the conclusion that angles E and F must sum to 60 degrees, while angles G and H at vertex B sum to 120 degrees. Thales' theorem is suggested as a potential solution, indicating that if AC is a diameter, then angle B is a right angle. The original poster realizes their confusion stemmed from mislabeling the angles, which complicated their calculations. Ultimately, the conversation highlights the importance of correctly identifying relationships between angles in geometric figures.
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I have the above equation to try and complete and I have no idea how to do it I have worked out all the angles for A, B, C, and D accept the ones within the diagonal line E and F. Any hints?
 

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I don't see an equation you are trying to complete, but what do we know about the opposite interior angles of a cyclic quadrilateral?
 
Sorry don't have an equation, my wording mistake, just trying to calculate the angle of E and F. What I have worked out is the angles of A, B, C, and D. A=70 B=110 C=120 D=60. But have no idea how to calculate E and F. If I turn C and D into a right angle on the outside of the trapezoid I get 30 and if I turn B and A into a right angle on the outside of the trapezoid I get 20. But not sure how to compute this into my E and F. Also I forgot to mention that A and D are parallel lines to B and C. Forgot to draw in the arrows on the line.Thank you for any help in solving this.
 
Since $\overline{AD}\parallel\overline{BC}$, and the trapezoid is cyclic, we know it must be isosceles. So we know $\angle A=60^{\circ}$, if $\measuredangle C=120^{\circ}$.

Thus, we know:

$$\angle E+\angle F=60^{\circ}$$

If we label the two angles at vertex $B$ as $G$ and $H$, we then know:

$$\angle G+\angle H=120^{\circ}$$

Can you find two more equations involving these 4 angles? Hint: add the interior angles of the two triangles making up the trapezoid...:D

Also, to simplify matters, $G$ and $H$ must share a special relationship to $E$ and $F$. ;)
 
[QUOTE Since $\overline{AD}\parallel\overline{BC}$, and the trapezoid is cyclic, we know it must be isosceles. So we know $\angle A=60^{\circ}$, if $\measuredangle C=120^{\circ}$.

I had changed the original degrees and didn't realize I had changed the triangles to equal sides. My triangles are not level only the horizontal lines are level, which leaves me with one obtuse angle and one acute angle. :S so sorry I confused matters, but I didn't want to get into trouble for my question from my teachers. Or am I confused lol.
 
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Hi lornick! Welcome to MHB! :)

Thales' theorem may help you.
200px-Thales%27_Theorem_Simple.svg.png

Thales' theorem: if AC is a diameter, then the angle at B is a right angle.
 
I like Serena said:
Hi lornick! Welcome to MHB! :)

Thales' theorem may help you.

Thales' theorem: if AC is a diameter, then the angle at B is a right angle.

Oh nooo I can't believe it was that simple, I have been trying to calculate it as a acute and obtuse angle (hence my issues). The time when you are doing maths and you start to bash your head against a brick wall, only to realize that the answer was right in front of you lol. Thank you sooooooooo much!
 
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