For $D$ on side $BC$ of $\triangle ABC$, with $K$ and $L$ circumcenters of $\triangle ABD$ and $\triangle ADC$, show $\triangle ABC\sim \triangle AKL$

Let $$D$$ be a point on side $$BC$$ of $$\triangle ABC$$. Let $$K$$ and $$L$$ be the circumcentres of $$\triangle ABD$$ and $$\triangle ADC$$, respectively.

Prove that $$\triangle ABC$$ and $$\triangle AKL$$ are similar.

Can I get a small hint on how to go start?

My attempts.

BL is a straight line and and ABDL is a kite. I think it can be proved by AAA. I've got that side AL is a diameter of circle AKL but don't know how to link it to angles A,B,C.

• In any case, show please your attempts. – Michael Rozenberg Apr 8 '20 at 1:54
• BL is a straight line and and ABDL is a kite. I think it can be proved by AAA. I've got that side AL is a diameter of circle AKL but don't know how to link it to angles A,B,C. – user377742 Apr 8 '20 at 2:27

$$KALD$$ is kite.

Thus,

$$\measuredangle ALK=\frac{1}{2}\measuredangle ALD=\measuredangle ACB,$$ $$\measuredangle AKL=\frac{1}{2}\measuredangle AKD=\measuredangle ABC$$ and we are done!

• +1. This deserved a diagram, so I took the liberty ... – Blue Apr 8 '20 at 6:22

Let $$R_B$$ and $$R_C$$ be the circumradii of the triangles ABD and ACD, respectively. Apply the sine rule to these two triangles,

$$AD = 2R_B\sin B = 2R_C \sin C$$

$$\frac{\sin B}{\sin C} = \frac{R_C}{R_B}=\frac bc\tag 1$$
So, the isosceles triangles ABK and ACL are similar and $$\angle BAK = \angle CAL$$, which yields
$$\angle BAC = \angle KAL$$