use vector, we need to prove: $\overrightarrow{AE}.\overrightarrow{FE}=0$ Note that: - $2\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{AQ}=\leftrightarrow (\overrightarrow{AB}+(1-k)\overrightarrow{AC})$, where $CQ=k.CA$ - $2\overrightarrow{FE}=\overrightarrow{FQ}+\overrightarrow{FB}=\overrightarrow{FQ}+\overrightarrow{FP}+\overrightarrow{PB}=2\overrightarrow{FD}+\overrightarrow{PB}$, where $D$ is midpoint of $PQ$

We shall admit the following elementary propositions: 1. A line bisecting two sides of a triangle is parallel to and is half as long as the third side; 2. If two triangles have one angle in the one equal to one angle in the other and if the sides of the equal angles are in proportion the two triangles are similar. Let triangle ABC as described; draw P, Q, circle F, QB, E, AE, and FE. Through F extend radius QF meeting circle at K; draw BK, and from E extend AE to meet BC and BK at R and N respectively. From A drop perpendicular AM meeting BC at M, and draw ME. To prove: AEF is a right angle. Notice that, in triangle QKB, FE bisects QB and QK; it follows that KB is parallel to FE [1], and therefore angle RNK is congruent to angle AEF. Also QK is perpendicular to BC for, arcs QPK and QCK being 180°arcs each, and being arcs QP and QC congruent for subtending congruent angles QCP and QPC respectively, it follows that arcs PK and CK are congruent and hence so will angles PQK and CPK; therefore QK is angle bisector of the vertex angle of an isosceles triangle and therefore is also altitude and median. Claim: Triangles CKB and MEA are similar. Indeed, ME bisecting both BC and BQ in triangle BCQ, it follows that ME is parallel to and is half as long as CQ [1], and hence angle AME is congruent to angle MAC, itself congruent with angle KQC, or KQP, which intercepts on circle F the same arc as angle PCK; therefore, angles AME and BCK are congruent. On the other hand: CK:CB :: CK:(2CM) ::(1/2)CK:CM; and being triangles QCK and AMC similar, we have: (1/2)CK:CM :: (1/2)QC:AM :: ME:AM; therefore, CK:CB :: ME:AM(*), hence triangles CKB and MEA have one angle in the one (BCK) congruent to one angle in the other (AME) and the sides of these angles proportional (*), and are therefore similar [2]. From this result, angles NBR and MAR are congruent and therefore triangles NBR and MAR have two angles congruent to two angles (angle MAR congruent to angle NBR, angle ARM to angle BRN as vertically opposed angles); therefore, their third angles are congruent; that is, BNR is congruent to right angle AMR; hence, RNK, that is, AEF is a right angle. Q.E.D.