Is it P4?

Davrbek wrote: Is it P4? Yep

Thanks))

FrenchFries356 wrote: Triangle $ABC$ with $AC=BC$ given and point $D$ is chosen on the side $AC$. $S1$ is a circle that touches $AD$ and extensions of $AB$ and $BD$ with radius $R$ and center $O_1$. $S2$ is a circle that touches $CD$ and extensions of $BC$ and $BD$ with radius $2R$ and center $O_2$. Let $F$ be intersection of the extension of $AB$ and tangent at $O_2$ to circumference of $BO_1O_2$. Prove that $FO_1=O_1O_2$. Let $O_1O_2$ and $AC$ intersection at $X$. And we need to prove that $\angle O_2FX=90$...... it is easy.

Mathuzb wrote: Let $O_1O_2$ and $AC$ intersection at $X$. And we need to prove that $\angle O_2FX=90$...... it is easy. Can you please provide your full solution?

Are points F, $O_1$ , D collinear?

No, they are not

It took too long. Let $O_1O_2\cap AC=T$. Let $F'$ be on the extension of $BA$ such that $F'O_1=O_1O_2=TO_1$. Because $AO_1$ is the external bisector of $TAF'$, we have that $TAO_1F'$ is cyclic, thus $\angle{TO_1F'}=2\angle{O_1O_2F'}=\angle{BAC}$, but $\angle{O_1BO_2}=\angle{BAC}/2$.

Mathuzb wrote: And we need to prove that $\angle O_2FX=90$...... it is easy. Could you prove please?

Suppose that the circle centered at $O_2$ is tangent to $AC$ at $L$ Let $O_1O_2$ meet $AC$ at a point $E$. Easy to see that $O_2LE$ is centered at $O_1$ Let $G$ be the unique point such that $GO_1=O_1O_2$ and $GEAO_1$ is convex. In directed angles: $$\angle EAO_1=\angle CAO_1=\angle O_1AG=\angle O_1EG=\angle EGO_1$$It follows that $GEAO_1$ is cyclic $$\angle GO_2E=\frac{1}{2}\angle GO_1E=\frac{1}{2}\angle GAE=\frac{1}{2}\angle BAC=\frac{1}{2}\angle CBA=\angle O_2BO_1$$From above we conclude that $GO_2$ is tangent to $BO_2O_1$ and $G=F$ which completes the proof.

Why, angle (O1BO2) = angle (ABC/2)??

By two angle bisectors