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.
Best Answer
Let $R_B$ and $R_C$ be the circumradii of the triangles ABD and ACD, respectively. Apply the sine rule to these two triangles,
$$AD = 2R_B\sin B = 2R_C \sin C$$
which leads to
$$\frac{\sin B}{\sin C} = \frac{R_C}{R_B}=\frac bc\tag 1$$
So, the isosceles triangles ABK and ACL are similar and $\angle BAK = \angle CAL$, which yields
$$\angle BAC = \angle KAL$$
Together with (1) for the side ratios, the triangle ABC and AKL are similar.