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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.

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    $\begingroup$ In any case, show please your attempts. $\endgroup$ – Michael Rozenberg Apr 8 '20 at 1:54
  • $\begingroup$ 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. $\endgroup$ – user377742 Apr 8 '20 at 2:27
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enter image description here

$KALD$ 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!

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    $\begingroup$ +1. This deserved a diagram, so I took the liberty ... $\endgroup$ – Blue Apr 8 '20 at 6:22
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enter image description here

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.

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