Hi emakarov,
My trig calculations are based on assuming that an 80-80-20 triangle exists as well as the 80 angle bisector. I show that the condition AD= BD +BC is true.And yes you found an error. It is cos80 times 10 not asprinted.
bjh
Begin with triangle DBC.
Imagine the rest of the figure is not yet in place.
Next attach triangle BQD which is congruent to triangle DBC.
Then attach triangle PQD which is also congruent.
The only way for P, D, C to be collinear is if $\displaystyle \alpha=60^o$
Finally attach triangle APQ.
We want to ensure that A, Q, B are also collinear if $\displaystyle \alpha=60^o$
and also that A, P, D, C are collinear.
$\displaystyle |\angle\ QPD|=40^o$
$\displaystyle |\angle\ DQP|=\omega$
$\displaystyle \alpha=60^o$
From triangle DBC, $\displaystyle \omega+\alpha+40^o=180^o\Rightarrow\ |\angle\ DQA|=\alpha+40^o$ if A, Q, B are collinear.
Then $\displaystyle |\angle\ APQ|=140^o$ if A, P, C, D are collinear.
From this it follows that $\displaystyle |\angle\ QAP|=20^o$
in order to fulfill the condition that A, P, D, C are collinear
and A, Q, B are collinear.
$\displaystyle \alpha=60^o$ creates both sets of collinearities.
It seems that your reasoning is as follows. If A, P, D, C are collinear, then $\displaystyle \angle BAC = 20$. Also, $\displaystyle \alpha=60$ iff A, P, D, C are collinear.
This only proves that if $\displaystyle \alpha=60$, then $\displaystyle \angle BAC=20$. Actually, this fact is obvious even with no additional construction beyond the problem statement: if $\displaystyle \alpha=60$, then $\displaystyle \omega=80$ from $\displaystyle \triangle BDC$, so $\displaystyle \angle BAC=180-2\cdot80=20$. I am not sure how to remove the assumption $\displaystyle \alpha=60$.
If we draw the lines $\displaystyle DQ,\;\;A'C',\;\;C'P$ as shown,
then extend a line from P so that $\displaystyle |\angle\ DPQ|=40^o$ this line will meet $\displaystyle A'$ but will it meet $\displaystyle Q\;\;?$
If the triangle $\displaystyle A'PC'$ is rotated about $\displaystyle D$ until $\displaystyle C'$ meets $\displaystyle C$ and $\displaystyle A'$ meets $\displaystyle A$
then "flip" it about $\displaystyle AC$ so that $\displaystyle P$ meets $\displaystyle B$
we will have both triangles completely superimposed on each other with $\displaystyle QD$ common if $\displaystyle \alpha=60^o$
It seems that you are proving that $\displaystyle \angle DPQ=40^\circ$. How does it help to prove that $\displaystyle \angle A=20^\circ$? Also, you write "... if $\displaystyle \alpha=60^\circ$". I agree that assuming $\displaystyle \alpha=60^\circ$ implies immediately that $\displaystyle \angle A=20^\circ$ and thus solves the problem. I just don't see how to prove that $\displaystyle \alpha=60^\circ$.
In that last one, I'm deliberately sending the line away from P in the direction of A' at 40 degrees.
It will definately meet A'.
What is being examined is the condition for which it goes through Q also.
The logic I'm using is that "if" alpha is 60 degrees, then both large triangles are identical,
hence if alpha is not 60 degrees, they will be different.
Therefore alpha needs to be 60 degrees.
Showing that the large triangles ABC and A'PC' are identical does not require the assumption that $\displaystyle \alpha=60$. They are equal because PC' = BC and the angles adjacent to those sides are equal.
Sorry, why? Obviously, a statement does not necessarily imply its inverse...The logic I'm using is that "if" alpha is 60 degrees, then both large triangles are identical,
hence if alpha is not 60 degrees, they will be different.
I've been trying a purely geometric proof. (wish my images would come out a bit smaller)
My first method was to ensure points were collinear.
I was hoping my 2nd attempt might be more convincing.
It hinges around angle QDP and the line DQ.
In the original orientation, it is 180-2(alpha) degrees.
In the A'C'P triangle, it becomes the middle of the three angles along a line.
The line DQ has the same orientation in both triangles.
Since the large external triangles are congruent without reference to Q being on the line A'P,
it's seen that Q will be on A'P only if alpha is 60 degrees,
allowing "internal" congruency also.
I will have another look to see if I can come up with another view,
though I hadn't intended any trigonometric analysis.
Thanks for being challenging!