Question 1209450
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A triangle ABC, where |AB| = |AC|, a line CD is drawn from angle C and intersects side AB at D, 
such that |AD| = |CD| = |BC|. Find the measure of angle A in degrees.
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            The solution and the answer  (72 degrees)  in the post by  @PChill is incorrect 


            I came to bring a correct solution.



<pre>
Draw triangle ABC with |AB| = |AC| and everything else from the problem on the sheet of paper.


Let x be the measure of the angle A;

    y be the measure of the angle B.



Consider triangle ABC. It is isosceles, since |AB| = |AC|  (given).
                       
                       Its "base" angle is y;  its "vertex" angle is x.

                       Therefore, x + 2y = 180°.    (1)



Consider triangle ADC. It is isosceles, since |AD| = |CD|  (given).

                       Its "base" angle is x;  its "vertex" angle is 180°-y   (from triangle BCD, which also is isosceles).

                       Therefore, 2x  + (180°-y) = 180°,

                       which implies  2x = y.       (2)


Now we have the system of two equations (1) and (2).

From (2), substitute y = 2x into equation (1).  You will get

    x + 2*(2x) = 180,

    x + 4x = 180,

      5x   = 180,

       x   = 180/5 = 36 degrees.


At this point, the problem is solved completely.


<U>ANSER</U>.  Angle A is 36 degrees.
</pre>

Solved.



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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Hello, &nbsp;@math_tuitor2020, &nbsp;please remove all your insinuations 

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