Let I be the incenter of ΔA₀A₁A₂.

For i = 0, 1, and 2:

   Let M_i be the midpoint of the side opposite vertex A_i.

   Let S_i be the reflection of M_i about ray A_iI.

Prove that rays A₀S₀, A₁S₁, and A₂S₂, are concurrent.
  

Writing A₀, A₁, A₂, M₀ as A, B, C, M, pro tem, and denoting

the intersections of A₀I and A₀S₀ with BC by P and Q, and the

lengths MP and PQ by d and e; then using usual notation:

Since AP bisects /QAM:   AM/AQ = d/e                 (1)

Since /MAC = /QAB, the sine rule used in triangles MAC and

QAB gives:        AM/AQ = (MC sinC)/(QB sinB) = (MC/QB)(c/b)

then, using (1):              d = e(MC*c)/(b*QB)

                                    = e(a/2)c/(b(a/2 – d – e))

from which:                   e = bd(a – 2d)/(ac + 2bd)           (2)

Since AP bisects /BAC:    PC/BQ = (a/2 + d)/(a/2 – d) = b/c

from which:                   d = a(b – c)/(2(b + c))                (3)

Using (3) to substitute for d in (2), after much simplifying
gives
                        e = abc(b – c)/((b + c)(b² + c²)

so that  BQ = a/2 – d – e   then simplifies to give

            BQ = ac²/(b² + c²)   and   CQ = ab²/(b² + c²)

Thus     BQ/CQ = c²/b²

Now reverting to the given notation, with Q becoming Q₀,

and having corresponding points Q₁ and Q₂ on other sides,

it follows that (using scalar lengths):

(A₁Q₀/A₂Q₀)(A₂Q₁/A₀Q₁)(A₀Q₂/A₁Q₂)  = (c²/b²)(a²/c²)(b²/a²) = 1

and, by the converse of Ceva’s theorem, A_iS_i, (i = 1,2,3) are

concurrent.

….there must be a simpler way! – should  BQ/CQ = c²/b²

be more readily obtainable??

Posted by Harry on 2016-08-24 17:03:58