Concurrency (I Think Using Menelaus' Theorem) Let $ABC$ be a triangle with incenter $I$, and let $B'$ and $C'$ be points on $BC$ such that $\angle{BIB'} = \angle{CIC'} = 90^\circ$. Let $AB'$ meet $CI$ at $P$, and let $AC'$ meet $BI$ at $Q$. Prove that $PQ$, $BC$, and the tangent to $\odot(BIC)$ at $I$ are concurrent.
So I have been trying to solve this problem by using Menelaus' theorem on $\triangle{BIC}$. So far, I have that if we the tangent at $I$ meet $BC$ at $R$, then by Menelaus theorem, we need $\frac{RB}{RC} \cdot \frac{BQ}{IQ} \cdot \frac{IP}{CP} = -1$. When trying to reduce this expression, the best I came up with was $\frac{RB}{RC} = \frac{IB^2}{IC^2}$. For those who are wondering, this result is not too hard to prove and can be done so using $\triangle{RBI} \sim \triangle{RIC}$. Any help on how to finish the problem would be very much appreciated. In particular, I am not sure how to deal with such seemingly nasty ratios. 
Sincerely, 
tworigami
 A: 
Probably there is an easier way to do it but my solution is as follows using conditional proof:
For this purpose, we can make a construction by connecting $BP$ such that it intersects the tangent to the circumcircle at $F$.
Let us assume that the three lines $PQ$, $BC$ and $IF$ are concurrent and the point of concurrency is $R$.
Like you suggest, applying Menelaus's theorem to $\triangle BIC$ and the line segment $RQ$, we have (you made a slight mistake in the ratios): $$\frac{CR}{BR} \cdot \frac{IP}{CP} \cdot \frac{BQ}{IQ} = -1$$
Similarly, applying the same theorem to $\triangle CBP$ and the line segment $RI$ yields: $$\frac{CR}{BR} \cdot \frac{IP}{IC} \cdot \frac{BF}{PF} = -1$$
Equating these ratios gives:$$\frac{IP}{IC} \cdot \frac{BF}{PF} = \frac{IP}{CP} \cdot \frac{BQ}{IQ}$$ $$\implies \frac{IC}{CP} \cdot \frac{PF}{BF} \cdot \frac{BQ}{IQ} = 1$$
which according to inverse Ceva's theorem on $\triangle IPB$ results in the concurrency between lines $IF$, $PQ$ and $BC$ asserting that our assumption is indeed correct. 
A: Can be done with barycentric coordinates. Wlog, let $BC + AC + AB = a + b + c = 1$. The condition for the orthogonality of $P_1P_2$ and $P_3P_4$ is that the permanent of the matrix with the rows $(a^2, b^2, c^2), P_2 - P_1, P_4 - P_3$ is zero. For $IB$ and $IB'$, this becomes
$$\operatorname{perm} \begin{pmatrix}
a^2 & b^2 & c^2 \\ a & b - 1 & c \\ a & b - y_{B'} & c - z_{B'}
\end{pmatrix} = 0 \;\Rightarrow\;
B' = (0, 1 - 2 c, 2 c).$$
Similarly, $C' = (0, 2 b, 1 - 2 b)$.
Then the intersection of $CI$ and $AB'$ is given by
$$\lambda_1 (0, 0, 1) + \mu_1 (a, b, c) =
\lambda_2 (1, 0, 0) + \mu_2 (0, 1- 2 c, 2 c) \;\Rightarrow \\
\frac {IP} {PC} = \frac {\lambda_1} {\mu_1} =
\frac {c (1 - 2 a)} {1 - 2 c}.$$
Replacing $c$ with $b$ gives $IQ/QB = b (1 - 2 a)/(1 - 2 b)$.
It's easy to show that the tangent to the circumcircle of $IBC$ is orthogonal to $IA$. Writing the orthogonality condition for $IA$ and $IR$, we get
$$\operatorname{perm} \begin{pmatrix}
a^2 & b^2 & c^2 \\ a - 1 & b & c \\ a & b - y_{R} & c - z_{R}
\end{pmatrix} = 0 \;\Rightarrow\;
\frac {BR} {CR} = - \frac {c (1 - 2 b)} {b (1 - 2 c)}.$$
Now (segments aren't signed, but the ratios are)
$$\frac {BR} {CR} \frac {CP} {IP} \frac {IQ} {BQ} =
-\frac {c (1 - 2 b)} {b (1 - 2 c)}
 \frac {1 - 2 c} {c (1 - 2 a)}
 \frac {b (1 - 2 a)} {1 - 2 b} = -1,$$
therefore $PQ$ goes through $R$.
