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I am solving a geometry problem as below enter image description here

I would like to find the value of $A'CA$.
I came up with three constructions, but they seem to be in short of one condition, namely:

  1. Construct $CD$ such that $CD\parallel AA'$ enter image description here But this doesn't show that $A'B\parallel AC$

  2. Reflect $\triangle A'AB$, but that doesn't show that $B'C$ is a straight line enter image description here

  3. Construct $BD$ such that $BD\parallel A'A$, but that doesn't show that $A'B\parallel AC$ as well. enter image description here

Can these constructions find the answer or there exists some restrictions and is this a coincidence?

xxxx036
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    I wouldn't use the word "solutions" but "constructions". – Jean Marie Oct 23 '21 at 08:30
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    Your constructions are short of one condition because I think you stopped too early. Unless there is some magic construction with just one or two lines, I think it requires you to draw some more lines and angles carefully. – Math Lover Oct 23 '21 at 13:02

1 Answers1

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enter image description here

Draw $BD \parallel A'A$. Then you can easily show that,

$AB = A'B = BD = CD$ and $\angle BAD = 70^\circ$

Now draw $BE$ and $AE$ such that $\angle BAE = \angle ABE = 60^\circ$. So, $BE = BD = CD$.

As $DE$ bisects $BC$ and $BC$ bisects $\angle DBE$, we have $CE = CD = AE$.

Finally, $\angle ACE = 40^\circ \implies \angle A'CA = 20^\circ$.

Math Lover
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