# Find $x$ angle in triangle

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.

• thats trigonometry solution. not what im looking for – Andriy Khrystyanovich Mar 1 at 17:14
• @Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic. – Michael Rozenberg Mar 1 at 17:18
• Note: A trigonometric solution is offered in this question. – Blue Mar 1 at 17:25
• math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry. – Matteo Mar 1 at 17:39
• – Aretino Mar 1 at 18:00

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

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

And my second solution is as follow:

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

• Very nice +1...... – Aqua Mar 1 at 17:56

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, $$\measuredangle BDC=10^{\circ}+20^{\circ}=30^{\circ}.$$

• Nice as always +1 – Aqua Mar 1 at 17:57