# Trigonometry/Geometry triangle help

In right triangle $$ABC$$, ($$\angle BAC = 90$$), $$D$$ is found on side $$AC$$, $$BD$$ is the angle bisector of $$\angle ABC$$ and $$BC= \sqrt3 BD$$.

1. Find the value of $$\angle CBD$$.

2. With regards to the point $$E$$ (found on $$AC$$), establish whether the areas of the circles which circumscribe the triangles $$BDE$$ and $$BCE$$ are equivalent.

3. (Not connected to number 2): Given that $$CE=2DE$$, find the value of the angle $$\angle BEC$$.

Note: I tried to call $$\angle ABC =\alpha$$ and thereby bisect it. I found different trigonometric relationships, but I could not work out how to solve for either the values of the sides nor the angles.

• For (1), can you express $BD$ in terms of $AB$ and $\angle ABC$? How about $BC$ (also in terms of $AB$ and $\angle ABC$)? So what does $BC=\sqrt{3}BD$ tell you? I'll leave (2) and (3) for later. Feb 15 at 15:52
• Sure, BD = AB divided by \sqrt3 cosine angle ABC Feb 15 at 15:56
• But I don't know how that helps me :( Feb 15 at 15:56

(1) By sine law,

$${\sin({\pi\over 2}-x)\over \sin({\pi\over 2}-2x)}=\sqrt3$$

$$\sin({\pi\over 2}-x)- \sqrt{3}\sin({\pi\over 2}-2x)=0$$

$$\cos(x)-\sqrt{3}\cos(2x)=0$$

$$\cos(x)-\sqrt{3}(2\cos^2(x)-1)=0$$

$$6\cos^2(x)-\sqrt{3}\cos(x)-3=0$$

$$\cos(x)={\sqrt{3}\pm\sqrt{75}\over 12}={\sqrt{3}\pm5\sqrt{3}\over 12}={1\over2}\sqrt{3},-{1\over3}\sqrt{3}$$

Since $$0 degrees, we take the positive result and $$x$$ is therefore $$30$$ degrees.

(2) $$BC>BE>BD$$ as given.

Angle $$C$$ correspond to a larger arc in the circumcircle of $$\triangle BCE$$ than in $$\triangle BCD$$. This means the circumcircle of $$\triangle BCE$$ is larger than the circumcircle of $$\triangle BCD$$.

Angle $$BDC$$ correspond to a larger arc in the circumcircle of $$\triangle BCD$$ than in $$\triangle BDE$$. This means the circumcircle of $$\triangle BCD$$ is larger than the circumcircle of $$\triangle BDE$$.

Combining both results, the circumcircle of $$\triangle BCE$$ is larger than the circumcircle of $$\triangle BDE$$.

(3) From (1) we already knows $$\triangle ABC$$ is a $$30,60,90$$ triangle. This means $$CD=2AD$$ therefore $$CD:DE:AD=4:2:3$$. Since $$AC:AB=\sqrt{3}$$ we know $$AE:AB={5\over 9}\sqrt{3}$$ and $$AB:AE={3\over 5}\sqrt{3}$$. Therefore $$\angle AEB=\arctan({3\over 5}\sqrt{3})$$ and $$\angle BEC=\pi - \arctan({3\over 5}\sqrt{3})$$

• I think dear friend that you should edit part 2 of your answer. Let S, A and A’ be the areas of the triangles BDC, BDE and BEC respectively. The point E starting from D to C, the area A increases from 0 to S while the area A' decreases from S to 0. This is a continuous process in which we will have in certain moment A = abc / 4R and A' = cde / 4R with the two equal radii R which is achieved when $\dfrac{ab}{A} =\dfrac{de}{A'}$. In such a case of equal radii the circumscribed circles are obviously of equal area. Feb 15 at 22:18
• @Piquito Why do you think the case you are describing is possible? When $A$ is $0$, $DE$ is also $0$ so it's a $0$ over $0$ thing and it's not obvious at all this is obtainable. In fact from my argument this is not obtainable. Feb 16 at 3:09
• @cr001.-I have no time dear friend by now. Maybe you are right but if the proportion i give $ab/A=de/A'$ is posible for certains values os sides and areas then you are wrong. Regards. Feb 17 at 2:47

1.$$AB=BD\cos(\alpha)\hspace 1cmAC=BC\sin(2\alpha)=\sqrt3BD\sin(2\alpha)\hspace 1cm BC=\sqrt3 BD$$

this implies $$2\sqrt3\cos(\alpha)^2-\cos(\alpha)-\sqrt3=0\Rightarrow \cos(\alpha)=\dfrac{\sqrt3}{2}$$. Then $$\angle{CBD}=30^{\circ}$$

2.The two respective radius should be equal. From the two areas $$A=\frac{BD\cdot DE\cdot BE}{4R}\\A'=\frac{BE\cdot EC\cdot BC}{4R'}$$ we have the condition $$R=R'\iff(BD\cdot DE)A'=(BC\cdot EC)A$$ This can be set straightforward but it is some tedious.

1. Note that $$\triangle{BAD}$$ is the half of an equilateral triangle so $$\angle{BDC}=120^{\circ}$$ and $$\angle{BCD}=30^{\circ}$$, etc.

You can take my answer as a simple comment if you consider it as incomplete.