Here's the elegant (in my opinion) inversive/projective solution instead . We want to prove that $AP=\sqrt 2$, or, in other words, $P$ is the antipode of $A$ in $\mathcal C=(ABC)$. Invert wrt the circle centered at $A$ of radius $1$. In general, we'll denote the image of $X$ by $X'$. The situation is the following one: We take points $E',F'$ on $\mathcal C$. Let $M'=E'D'\cap BF$, and $N'=E'C\cap AF'$. We want to show that $M'N'\cap BC$ is precisely $D$ (the situation is a bit reverted, but this is easily seen to be equivalent to the fact that $P=D'$, which is the antipode of $A$ in $\mathcal C$). Fix $E'\in\mathcal C$ and move $F'$ on the same circle. The map $N'\mapsto M'$ that this movement induces from the line $E'C$ to the line $E'D'$ is clearly projective, and it's also clear that $E'=E'C\cap E'D'$ corresponds to itself through this transformation, which means that the transformation is a perspectivity, i.e. $N'M'$ passes through a fixed point. This point must be equal to the intersecton of the lines $N'M'$ which we get when we make $F'=D',F'=C$, and this point is $D'$.

I should ask for a non-inversive and non-projective solution Anyway, I'm not very familiar with projective geometry, but indeed a cool proof Actually what I'm expecting is a solution by radical axis which I believe is possible. btw, the solution in my book used the same inversion and followed by Pascal.

For now, i can only say in my geometers sketchpad's opinion, for any isosceles triangle $AP$ is the diameter of circumcircle of $ \bigtriangleup ABC$.

xeroxia wrote: for any isosceles triangle $AP$ is the diameter of circumcircle of $ \bigtriangleup ABC$. Indeed, we have: Extended problem. Let ABC be an isosceles triangle with AB = AC, and let D be the midpoint of its base BC. Let E and F be two points on the line BC. Let the circles ADE and ABF meet at a point M (apart from A), let the line AF meet the circle ACE at a point N (apart from A). Finally, let the line AD meet the circle ABC at a point Q (apart from A). Then, prove that the segment AQ is a diameter of the circumcircle of triangle ABC, and that the point Q lies on the circle AMN. Hereby, "circle $P_1P_2P_3$" means the circle through three given points $P_1$, $P_2$, $P_3$. Of course, the extended problem yields that Q = P, and thus your assertion is proven. So let's solve the extended problem: Since the triangle ABC is isosceles with AB = AC, the line AD is its symmetry axis, and thus, the point Q where this line meets the circumcircle of triangle ABC (apart from A) is the point diametrically opposite to A on this circumcircle. In other words, the segment AQ is a diameter of the circumcircle of triangle ABC. What remains to prove is that the point Q lies on the circle AMN. Now, the circle with center A and radius AB = AC passes through the points B and C. Thus, the inversion with respect to this circle leaves the points B and C fixed. Hence, the image of the line BC under this inversion is the circle ABC (in fact, the image of the line BC under our inversion is a circle through the inversion center A, and passing through the images of the points B and C, i. e. these points B and C themselves, so it must be the circle ABC). On the other hand, the inversion leaves the line AD fixed (like every line through the inversion center A). Thus, the point D, where the line AD meets the line BC, is mapped to the point Q, where the line AD meets the circle ABC. Since our inversion maps the line BC to the circle ABC, the points E and F, lying on the line BC, are mapped to two points E' and F' on the circle ABC. The circles ADE, ABF, ACE and ABC are mapped to the lines QE', BF', CE' and BC, respectively (in fact, all circles passing through the inversion center A get mapped to lines, and these lines pass through the images of the respective points on the circles). The line AF remains fixed (as any line through the inversion center A does), but we could also say that it is mapped to the line AF' (in fact, the lines AF and AF' are the same, since the point F' is the image of F under our inversion with center A, and thus lies on the line AF). Hence, the point of intersection M of the circles ADE and ABF is mapped to the point of intersection M' of the lines QE' and BF', and the point of intersection N of the line AF with the circle ACE is mapped to the point of intersection N' of the lines AF' and CE'. Finally, the inversion maps the circle AMN to the line M'N' (again, for the same reasons: the circle passes through the inversion center A, so it becomes a line, and this line passes through the images of the points of the circle). And, of course, the point Q is inverted into the point D (since Q is the image of D, and inversion is an involution). Thus, the assertion we have to prove, namely the assertion that the point Q lies on the circle AMN, is equivalent to the assertion that the point D lies on the line M'N'. But this is easy to prove: The hexagon BF'AQE'C is inscribed (in fact, all of its vertices lie on the circle ABC), and thus, by Pascal's theorem, the points $BF^{\prime}\cap QE^{\prime}$, $F^{\prime}A\cap E^{\prime}C$ and $AQ\cap CB$ are collinear. But $BF^{\prime}\cap QE^{\prime}=M^{\prime}$, $F^{\prime}A\cap E^{\prime}C=N^{\prime}$ and $AQ\cap CB=D$. Hence, the points M', N' and D are collinear, i. e. the point D lies on the line M'N'. And the problem is solved. Personally, I don't like problems which are just obtained by inverting some well-known configuration... darij