$\triangle ABC$ with circumcenter $K$ has $AB=AC$ and $AD=BC$. Find $\angle BAC$.

Question

In a triangle, $AB = AC$ and $D$ is its circumcenter. It is also known that $AD = BC$. Find the measure of the $\angle BAC$.

Any tips for me being able to resolve the issue? I tried to let k triangle inscribed.

Answer 1

Assuming D is your circumcenter (and that there is a typo in your question)

Check that $AD=DB=DC=BC$

So first $\triangle DBC$ is equilateral. So angle $DBC=$ angle DCB $=60^{\circ}$

Also check that $\triangle ABD$ and $\triangle ADC$ are isosceles. So angles $ABD=BAD=DAC=DCA=\alpha$(say)

Using that the sum of all angles in a triangle is $180^{\circ}$, we get angle $BAC=2\alpha=30^\circ$

Another solution.

Imagine a circle anround the $\triangle ABC$. Since D is the center of your circle, the segment BC will subtend half the angle it subtends on the centre as it subtends on the cicumference. (half of $60^\circ$ )

Answer 2

Let $\measuredangle BAC=\alpha$ and $R$ be a radius of the circle.

Thus, by the law of sines we obtain: $$R=2R\sin\alpha$$ or $$\sin\alpha=\frac{1}{2},$$ which gives $$\alpha=30^{\circ}$$ or $$\alpha=150^{\circ}$$

Answer 3

The answer of Rozenberg is exact with two cases, but we can solve this issue with primary geometry.We always have $\triangle BDC$ is equilateral. But depending on the position of $D$ which is inside or outside of $\triangle ABC$ (in fact, $\triangle ABC$ is obtuse or acute), we have two cases.

Case 1:

Case 2: