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In an isosceles triangle ABC, the angle of the apex vertex A equals to 20 degrees. Point D exists on AB so that AD=BC. Find the angle of BCD.

We must have to use the equality of AD and BC in someway. But I have no idea how to use it. I saw solutions using trigonometry but I want a non-trigonometry solution.

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    You say the triangle $ABC$ is isosceles. Is it know which two sides are equal? – 2017-02-17
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    AB and AC are equal. – 2017-02-17

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Construct equilateral triangle $ADE$ outside triangle $ABC$. Let $M$ be the midpoint of $AE$. Since $ADE$ is equilateral, $DM$ is the orthogonal bisector of segment $AE$. Moreover, since $AE = AD = BC$ and $$\angle \, EAC = \angle \, EAD + \angle \, BAC = 60^{\circ} + 20^{\circ} = 80^{\circ} = \angle \, BCA$$ triangles $CEA$ and $ABC$ are congruent. However, $ABC$ is isosceles so $CEA$ is also isosceles where $CA = CE$ and $\angle \, ACE = \angle \, CAB = 20^{\circ}$. Therefore point $C$ lies on the orthogonal bisector of segment $AE$, which we already proved is $DM$. Hence, the three points $M, D$ and $C$ are collinear. The line $CD$, which coincides with $DM$, is thus the angle bisector of $\angle \, ACE = 20^{\circ}$ so $\angle \, DCA = 10^{\circ}$. Therefore $$\angle \, BCD = \angle \, BCA - \angle \, DCA = 80^{\circ} - 10^{\circ} = 70^{\circ}$$

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    Sir how does one develop geometric intuition like you? Is it just practice? I am a fan of your synthetic proofs. – 2017-02-18
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    @N.S.JOHN Thank you. However, I am not sure I know the answer to your question and I cannot state with clear conscience that it is just practice. It was this kind of geometry in high school together with math's modeling power and applications to physics that turned me into a mathematician. But for sure, experience is of great help. – 2017-02-19