It's actually this same problem http://www.gogeometry.com/school-college/p894-triangle-incenter-60-degrees-midpoint-parallel-html5-ipad.htm, but first you have to prove that angle A = 60 degrees, the circumcircles of ABC and BHC have equal radii, and the two lines PC and BQ are parallel.

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We will prove 2 main claims. Claim 1: A, I, T are collinear if and only if $\angle BAC = 60$. Proof: Let the extension of AH, BH, CH meet BC, CA, AB at D, E, and F respectively. Since TB = TH = TC, we have $$\angle BTC = 2 (180 - \angle BHC) = 2 (180 - \angle BHF) = 2 (180 - \angle BFH - \angle ABE) = 2 (180 - \angle BFH - 180 + \angle AEB + \angle BAE) = 2 \angle BAC$$Thus we have $$\angle BAC = 60 \Longleftrightarrow \angle BAC + \angle BTC = 60 + 2 \cdot 60 = 180 \Longleftrightarrow BACT \text{ is a cyclic quadrilateral} \Longleftrightarrow A, I, T \text{ are collinear},$$where we can also deduce that T is the A-excenter. Claim 2: $\angle BAC = 60$ if and only if $[BKS] = [CKR].$ Proof: Let U, V be altitudes taken from I to AB and AC respectively. Obviously $IU = IV = r = \frac{2[ABC]}{a + b + c}$. We have $$[ASI] + [API] = [ASP] \Longleftrightarrow IU \cdot AS + IV \cdot AP = AS \cdot AP \cdot \sin{A}$$where using some simple calculations we obtain $AS = \frac{bc}{a+b-c}$ and $AR = \frac{bc}{a-b+c}$. Thus, $$[BKS] = [CKR] \Longleftrightarrow [ABC] = [ASR] \Longleftrightarrow bc = \frac{(bc)^2}{(a+b-c)(a-b+c)} \Longleftrightarrow a^2 = b^2 + c^2 - bc,$$which is obviously true by the LoC in triangle ABC.