Estonian Mathematical Olympiad

We first show that $\angle BGF=90^{\circ }$ (Fig. 17). For this we notice that $F$ as the midpoint of the hypothenuse in $CDE$ is also its circumcenter, so $FC=FD$. Together with $BEGC$ being cyclic, we get $$\angle EGB=\angle ECB=\angle FCD=\angle FDC.$$ Thus $$\begin{array}{rl}\angle BGF& =\angle EGF-\angle EGB\\ & =180^{\circ }-\angle EDF-\angle FDC=180^{\circ }-\angle EDC=180^{\circ }-90^{\circ }=90^{\circ }.\end{array}$$

Fig. 17 Fig. 18

Let $A^{\prime }$ and $K$ be the intersections of $GF$ with $AD$ and with the circumcircle of $ABC$, respectively (Fig. 18). As $\angle BGK=\angle BGF=90^{\circ }$, we have that $BK$ is a diameter of the circumcircle of $ABC$. Then also $\angle BCK=90^{\circ }$, from which $CK\parallel AE$. Now $AKCE$ as a cyclic quadrilateral with two parallel sides must be an isosceles trapezium. Combining this with $FD=FE$, we get $\angle EDF=\angle DEF=\angle AEC=\angle EAK$, so $DF\parallel AK$. From $FE=FC$ and the parallel lines we get equal triangles $FCK$ and $FE{A}^{\prime }$, from which $F{A}^{\prime }=FK$. Thus $DF$ is a midline of $AK{A}^{\prime }$, from which $D$ is the midpoint of $A{A}^{\prime }$. In conclusion $D$ is the circumcenter of the right-angled triangle $AP{A}^{\prime }$, which finishes the proof.

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Alternative solution.

Let $A^{\prime }$ be the reflection of $A$ over $BC$ and let $BK$ be a diameter of the circumcircle of $ABC$. First, we will show that $F$ lies on $A^{\prime }K$ (Fig. 19). As $KC\perp BC$ and $AD\perp BC$, we have $KC\parallel E{A}^{\prime }$. Also $$\begin{array}{rl}\angle AEK& =\angle ACK=90^{\circ }-\angle ACB,\\ \angle A{A}^{\prime }C& =\angle D{A}^{\prime }C=\angle DAC=90^{\circ }-\angle ACB.\end{array}$$ Thus $\angle AEK=\angle A{A}^{\prime }C$, from which $EK\parallel A^{\prime }C$. Therefore $A^{\prime }CKE$ is a parallelogram. As the intersection of the diagonals of a parallelogram divides both of its diagonals in half and $F$ is the midpoint of $CE$, it must also be the midpoint of $A^{\prime }K$, so $A^{\prime },K,F$ are indeed collinear.

Fig. 19

Now we show that $G$ also lies on $A^{\prime }K$. For this, let $G^{\prime }$ be the second intersection of $A^{\prime }K$ and the circumcircle of $ABC$ (Fig. 20). Notice that $DF$ is a midline of $A^{\prime }AK$, so $DF\parallel AK$. Therefore $$\angle EDF=\angle EAK=180^{\circ }-\angle E{G}^{\prime }K=180^{\circ }-\angle E{G}^{\prime }F,$$ which shows that $G^{\prime }$ lies on the circumcircle of $DEF$, meaning that $G^{\prime }=G$. We have shown that $A^{\prime }$ lies on $GF$, which also contains $P$. Thus $AP{A}^{\prime }$ is a right triangle, whose circumcenter is the midpoint $D$ of $A{A}^{\prime }$. Therefore $DA=DP$, as desired.

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Alternative solution.

Like in Solution 1, we show that $FC=FD$ and $BG\perp GF$. Since also $AP\perp GF$ (Fig. 21), we have $AP\parallel BG$. From this $$\angle GAP=\angle AGB=\angle ACB=\angle ACD,$$ so the right triangles $GAP$ and $ACD$ are similar. Therefore $$\frac{GA}{AC}=\frac{PG}{DA}.(7)$$

Fig. 21

On the other hand, we have the equalities $$\angle GDF=\angle GEF=\angle GEC=\angle GAC,$$ $$\angle FGD=\angle FED=\angle CEA=\angle CGA,$$ from which it follows that the triangles $GDF$ and $GAC$ are similar. Thus $$\frac{GA}{AC}=\frac{GD}{DF}.(8)$$

From (7) and (8) we get $\frac{PG}{DA}=\frac{GD}{DF}$ or $$\frac{GD}{PG}=\frac{DF}{DA}.(9)$$ Notice also that there must exist a spiral similarity at $G$ between the similar triangles $GDF$ and $GAC$. Hence $AGD$ and $CGF$ are also similar, implying $$\frac{GC}{GA}=\frac{FC}{DA}.(10)$$ As $DF=FC$, the right hand sides of (9) and (10) are equal. So the left hand sides must be equal as well, i.e., $\frac{GD}{PG}=\frac{GC}{GA}$. As $\angle PGD=\angle FGD=\angle CGA$, triangles $PGD$ and $AGC$ must also be similar. Therefore $$\frac{GA}{AC}=\frac{PG}{DP}.(11)$$ Combining (7) and (11) gives $DA=DP$, as desired.