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I need to find angle x in this isosceles triangle(20-80-80), by using pure geometry, if i can say so. If my calculations are correct (i tried another approach) answer should be 30, but there should be 'easy' way to find this. Also i found many Langley’s Adventitious Angles exercises which are very similar to mine but yet different.

enter image description here

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Construct an equilateral triangle such that its sides are equal to the base of the main triangle. enter image description here

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  • $\begingroup$ Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer. $\endgroup$ – Aqua Mar 1 at 17:54
  • $\begingroup$ @greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules. $\endgroup$ – Seyed Mar 1 at 18:02
  • $\begingroup$ Yes, that is correct if you want another upvote. $\endgroup$ – Aqua Mar 1 at 18:03
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    $\begingroup$ @greedoid, Thanks for your advise. $\endgroup$ – Seyed Mar 1 at 18:06
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And my second solution is as follow: enter image description here

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enter image description here

construct triangle $\Delta BCE$ congruent to $\Delta ADB$.

so $AB = BE$, $\angle ABE = 80° - 20° = 60°$

Thus triangle $\Delta ABE$ is equilateral.

$AB = AE = AC$, since $\angle CAE = 60° - 20° =40°$

$\angle AEC = \frac{180° - 40°}{2} = 70°$

so $x = 20° + \angle ABD = 20° + \angle CBE = 20° + (70° - 60° ) = 30°$

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  • $\begingroup$ Very nice +1...... $\endgroup$ – Aqua Mar 1 at 17:56
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Let in $\Delta ABC$ we have $AB=AC$, $\measuredangle A=20^{\circ}$ and $\measuredangle ADC=x$ as on your picture.

Let $M\in AB$ such that $AD=MD$ and $K\in DC$ such that $MK=AD$.

Also, let $B'\in MB$ such that $MB'=AD$ and $C'\in KC$ such that $B'C'||BC.$

Thus, $$\measuredangle MKA=\measuredangle MDK=2\cdot20^{\circ}=40^{\circ}$$ and from here $$\measuredangle B'MK=40^{\circ}+20^{\circ}=60^{\circ},$$ which says $$B'K=MB'=AD=BC.$$ But $$\measuredangle B'KC'=60^{\circ}+20^{\circ}=80^{\circ}=\measuredangle BCA=\measuredangle B'C'A.$$

Thus, $$B'C'=B'K=AD=BC,$$ which says that $$B\equiv B'$$ and $$C\equiv C'.$$ Id est, $$\measuredangle BDC=10^{\circ}+20^{\circ}=30^{\circ}.$$

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  • $\begingroup$ Nice as always +1 $\endgroup$ – Aqua Mar 1 at 17:57

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