Can't Solve Easy Geometry Problem

[RandomGuy1520]

Formalized:

a b d = triangle a b d; c = on_circle c a b; m = midpoint m b c; e = on_line a d, eqangle3 b a b a d; o = circle o b d c; u = midpoint u a d; k = on_circle k o b, on_circle k u d; r = circle r a m k ? cyclic a k m e

Example proof:

• Let KD and BC intersect at T and construct L on AK such that $AK \times AL=AB^2$.
• Since <BEA=<ABD, so $AD \times AE=AB^2=AK \times AL$ so DKLE cyclic.
• So E, L is the inverse of D, K with origin A and radius AB (=AC).
• Since BDKC is cyclic by definition of K, BELC is cyclic by inversion (or you can prove this using plain angle chasing).
• So <DEL=<AKD=pi/2
• By the radical axis theorem on circles BDKC, BELC, DKEL gives TEL is collinear.
• So <TEA=pi/2
• Since <TMA=<TKA=pi/2 by definition of T and K, so AKME is cyclic on the circle of diameter AT.

Btw you can also prove this using purely length chasing, here not shown.

[witchandcherry]

Alphageometry doesn't use power of point, so ye, it can't solve this problem