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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.
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.
ANSER. Angle A is 36 degrees.
Solved.
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Hello, @math_tuitor2020, please remove all your insinuations
about me from all your posts - WITHOUT DELAY.
Jan. 20, 2025.
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