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How to find the angles at trapezium $ABCD$ (that is $\angle DAB$, $\angle ABC$, $\angle BCD$ and $\angle CDA$) if known $\angle DAC=80^{\circ}$ and $AD=AB=BC$?

  • $AC$ - diagonal;

  • $AD$ and $BC$ - sides of the trapezium;

  • $AB$ and $DC$ - bases of the trapezium;

There is drawing

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1 Answer 1

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Hint:

1) $$AB||CD \Rightarrow \angle BAC= \angle DAC=x$$

2) $$AB=BC \Rightarrow \angle BAC= \angle BCA=x$$

3) $$AD=BC \Rightarrow \angle ADC= \angle DCB=2x$$

4) $$\angle ADC=\angle DAC=180^{\circ}=2x+80^{\circ}+x$$

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