Find $x$ angle in triangle

Question

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.

@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 01 '19 at 17:18
https://math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry. — dfnu, Mar 01 '19 at 17:39
Possible duplicate of In $\triangle ABC$ with $AB=AC$ and $\angle BAC=20^\circ$, $D$ is on $AC$, with $BC=AD$. Find $\angle DBC$. Where's my error? — Intelligenti pauca, Mar 01 '19 at 18:00

score 12 · Accepted Answer · edited Mar 01 '19 at 19:23

12

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

edited Mar 01 '19 at 19:23

Acccumulation

12,210

answered Mar 01 '19 at 17:45

Seyed

8,933

Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer. – nonuser Mar 01 '19 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 01 '19 at 18:02
Yes, that is correct if you want another upvote. – nonuser Mar 01 '19 at 18:03
2

@greedoid, Thanks for your advise. – Seyed Mar 01 '19 at 18:06

score 7 · Answer 2 · answered Mar 01 '19 at 18:04

7

And my second solution is as follow:

answered Mar 01 '19 at 18:04

Seyed

8,933

Michael Rozenberg · Answer 3 · 2019-03-01T17:35:52.683

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

score 5 · Answer 4 · edited Mar 01 '19 at 19:37

5

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

edited Mar 01 '19 at 19:37

Dr. Mathva

8,621

answered Mar 01 '19 at 17:53

qsmy

545

Very nice +1...... – nonuser Mar 01 '19 at 17:56

Find $x$ angle in triangle

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