Before proving these propositions, we shall prove the following lemma. Lemma. If two triangles on the same base and on same side of it have equal interior vertical angles, the extremities of the base and the vertices are concyclic. Proof. (see 1st atachment) Let $\triangle ABC$ and $\triangle ABD$ two triangles on same base $\overline {AB}$ and on same side of it so that $\angle D \cong \angle C$. About $\triangle ABC$ circumscribe a circle (from 3 points not on same line, one circle ad only one may be descried); then the circle passes through $D$. Indeed, suppose, if possible, that $D$ falls inside the circle, like on point $E$; then $\angle BEA \cong \angle C$. Through $E$, extend $\overline {BE}$ to meet the circle in $G$; then $\angle BEA$, being exterior to $\triangle EGA$, is larger than the opposite interior $\angle BGA$ (in a triangle, an exterior angle is larger than either opposite interior angles); but $\angle BGA \cong \angle C$; hence $\angle BEA > \angle C$, which is contrary to hypothesis that $\angle BEA \cong \angle C$; therefore $D$ could not fall inside the circle. Suppose now, if possible, that point $D$ falls outside the circle like on point $F$, hence, $\angle BFA \cong \angle C$. Now, interior $\angle BFA$ of $\triangle GFA$ is smaller than the triangle’s remote exterior $\angle BGA$; but $\angle BGA \cong \angle C$; hence $\angle BFA$ is smaller that $\angle C$, contrary to the hypothesis that they are congruent. Therefore $D$ could not fall outside the circle. It follows that $D$ could fall neither inside nor outside the circle; and hence it must fall on the circle: hence the extremities of the base and the two vertices are concyclic. Q.E.D.

Attachments:

Now back to the main problem: (See attachment) Let $\triangle ABC$ and $A_1, B_1, C_1$, either interior or exterior points on $\overline {BC}, \overline {CA}, \overline {AB}$, respectively, such that $\overline {AA_1}, \overline {BB_1}, \overline {CC_1}$ meet in $P$ and such that $A, B_1, A_1, B$ are concyclic and $B, C_1, B_1, C$ are concyclic. 1st case. $P$ is interior to $\triangle ABC$ as in the right figure. Join in succession $A_1. B_1, C_1$; then quadrilaterals $AB_1A_1B$ and $BC_1B_1C$ are cyclic. Hence, in quadrilateral $AB_1A_1B$, $\angle B_1AA_1 \cong \angle B_1BA_1 \cong \angle B_1BC$; but in quadrilateral $BC_1B_1C$, $\angle B_1BC \cong \angle B_1C_1C$; therefore, $\angle B_1AA_1\cong \angle B_1C_1C$. Now, consider $\triangle APB_1$ and $\triangle C_1PB_1$; they have a common base $\overline {PB_1}$, are on same side of it and have two equal vertex angles; that is $\angle B_1AP$ (or $\angle B_1AA_1$), and $\angle B_1C_1P$ (or $\angle B_1C_1C$); hence, according to the previous lemma, $A, B_1, P, C_1$ are concyclic and quadrilateral $AB_1PC_1$ is cyclic. It follows that $\angle PAC_1 \cong \angle A_1AC_1 \cong \angle PB_1C_1 \cong \angle BB_1C_1$; but $\angle BB_1C_1 \cong \angle BCC_1 \cong \angle A_1CC_1$; hence $\angle A_1AC_1 \cong \angle A_1CC_1$. Therefore, noticing $\triangle A_1CC_1$ and $\triangle A_1AC_1$, on same side of the common base $\overline {A_1C_1}$; they have congruent interior vertex angles, $\angle A_1AC_1$ and $\angle A_1CC_1$; therefore, according to lemma above, $A, C_1, A_1, C$ are concyclic. This proves point a). Now, for point b), notice that cyclic quadrilateral $BC_1B_1C$ and $AB_1A_1B$ share $\angle ABC$ and therefore their respective opposite interior angles are conguent (opposite angles of a cyclic quadrilateral are supplementary), and hence the supplementary of these equal angles must be congruent to each other and to $\angle ABC$ (an angle and the supplementary of its supplementary are congruent); that is, $\angle AB_1C_1 \cong \angle CB_1A_1 \cong \angle ABC$; but we have proved that $\angle C_1B_1B \cong \angle BB_1A_1$; and hence, $\angle AB_1B \cong \angle CB_1B$; since they are supplementary, each angle is a right angle. Therefore, $\angle AB_1C_1$ and $\angle C_1B_1B$ are complementary, but $\angle ABA_1 \cong \angle AB_1C_1$ and $\angle A_1AB \cong \angle BCC_1 \cong \angle C_1B_1B$; hence, $\angle ABA_1$ and $\angle BCC_1$ are both complementary to $\angle A_1AB$ and therefore, in $\triangle ABA_1$, $\angle AA_1B$ is a right angle; and in $\triangle CBC_1$, $\angle CC_1B$ is a right angle. 2nd case. $P$ is exterior to $\triangle ABC$ as in the left figure. Join in succession points $A_1, B_1, C_1$; then quadrilateral $BCC_1B_1$ is cyclic and shares with cyclic quadrilateral $AB_1BA_1$ interior $\angle PBC$: therefore, their opposite exterior angles are equal to each other and equal to $\angle PBC$; that is, $\angle B_1AP \cong \angle B_1C_1P \cong \angle PBC$; but then, $\triangle B_1PA$ and $\triangle B_1C_1P$, on same side of common base $\overline {B_1P}$, have equal interior vertical angles and therefore, according to previous lemma, $P, B_1, A, C_1$ are concyclic and quadrilateral $PB_1AC_1$ cyclic. Hence, $\angle PB_1C_1 \cong \angle PAC_1$ and $\angle PAC_1 \cong \angle BAA_1$ as vertically opposed angles; therefore, $\angle PB_1C_1 \cong \angle BAA_1$. Noticing that $\angle PB_1C_1$ is an exterior angle to cyclic quadrilateral $BCC_1B_1$, it follows that $\angle PB_1C_1 \cong \angle A_1CC_1$. Thus, $\angle BAA_1 \cong \angle A_1CC_1$. This means that an exterior angle ($\angle BAA_1$) and an opposite interior angle ($\angle A_1CC_1$) of quadrilateral $AC_1CA_1$ are congruent, which means that it is cyclic and its vertices concyclic. This proves point a). For point b), notice that $\angle BB_1A_1 \cong \angle BAA_1$ ($AB_1BA_1$ cyclic); hence considering quadrilateral $PCA_1B_1$, an exterior angle ($\angle BB_1A_1$) and an opposite interior angel ($\angle A_1CC_1$) are congruent and is therefore cyclic. Considering the left figure from the standpoint of $\triangle PBC$, with three line segments meeting in $A$, it is equivalent to the first case, for, two quadrilaterals having these lines as diagonals are cyclic: $BCC_1B_1$ by hypothesis, and $PCA_1B_1$ by proof; therefore the same conclusions may be drawn as in first case, that is, $PC_1A_1B$ is cyclic and that the angles made by the three line segments $\overline {PA_1}, \overline {BC_1}, \overline {CB_1}$ and the sides of $\triangle PBC$ are right angles from which it is plain that $\overline {AA_1}, \overline {BB_1}, \overline {CC_1}$ are respectivly perpendicular to $\overline {BC}, \overline {AC}, \overline {AB}$. Q.E.D.

Attachments:

This is another proof using Ceva's Theorem together with the lemma above: From the theorem of Ceva, we have: $\frac{AB_1 \cdot CA_1 \cdot BC_1}{B_1C \cdot A_1B \cdot C_1A}=1;$ but $AB_1A_1B$ being cyclic, $\triangle AB_1P \sim \triangle BA_1P$, and hence, $\frac{AB_1}{A_1B}=\frac{AP}{BP}=\frac{PB_1}{PA_1} (1)$; and from $BC_1B_1C$ being cyclic, $\triangle BC_1P \sim \triangle CB_1P$, and hence, $\frac{BC_1}{B_1C}=\frac{BP}{CP}=\frac{PC_1}{PB_1} (2)$; so that $\frac{AB_1 \cdot CA_1 \cdot BC_1}{B_1C \cdot A_1B \cdot C_1A}=\frac{AB_1 \cdot BC_1 \cdot CA_1}{A_1B \cdot B_1C \cdot C_1A}=\frac{AP}{BP} \cdot \frac{BP}{CP} \cdot \frac{CA_1}{C_1A}=1.$. It follows that $\frac{CA_1}{C_1A}=\frac{CP}{AP}. (3)$ On the other hand, $\frac{CP}{AP}=\frac{CP}{B_1P} \cdot \frac{PB_1}{AP}$; and by using (1) and (2) again we have $\frac{CP}{B_1P}=\frac{PB}{PC_1}$, and $\frac{PB_1}{AP}=\frac{PA_1}{PB}$, so that $\frac{CP}{AP}=\frac{PB}{PC_1} \cdot \frac{PA_1}{PB}$, that is $\frac{CP}{AP}=\frac{PA_1}{PC_1}. (4)$ It follows from (3) and (4) that $\frac{CA_1}{C_1A}=\frac{CP}{AP}=\frac{PA_1}{PC_1}$, which means that the sides of $\triangle AC_1P$ and $\triangle CA_1P$ are proportional and therefore, $\triangle AC_1P \sim \triangle CA_1P$, from which it follows that $\angle A_1AC_1 \cong \angle C_1CA_1$ and hence by applying lemma above on triangles, $\triangle A_1AC_1$ and $\triangle C_1CA_1$ with common base $\overline {A_1C_1}$ and on same side of it, we have that $AC_1A_1C$ is cyclic. This proves a). Now. from the quadrilaterals $AB_1A_1B$, $BC_1B_1C$ and $AC_1A_1C$ being cyclic and from the resulting similarity of the pairs of triangles $\triangle APC_1$ and $\triangle CPA_1$; $\triangle APB_1$ and $\triangle BPA_1$; $\triangle CPB_1$ and $\triangle BPC_1$, we have $\angle A_1AB \cong \angle C_1CB$, $\angle A_1AC \cong \angle B_1BC$, and $\angle C_1CA \cong \angle B_1BA$; from which it is easy to conclude that in triangles $\triangle A_1AB$ and $\triangle A_1AC$, the angles $\angle A_1AB$ and $\angle ABA_1$ together add to the same measure as the angles $\angle A_1AC$ and $\angle ACA_1$ together, making the supplementary angles $\angle AA_1B$ and $\angle AA_1C$ congruent which makes each a right angle. A similar reasoning leads to the same conclusion about the angles made at $B_1$ and $C_1$ respectively by $\overline {BB_1}$ and $\overline {AC}$ and by $\overline {CC_1}$ and $\overline {AB}$. Q.E.D.