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 Concurrent Rays (Posted on 2016-08-20)

Let I be the incenter of ΔA0A1A2.

For i = 0, 1, and 2:

Let Mi be the midpoint of the side opposite vertex Ai.

Let Si be the reflection of Mi about ray AiI.

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

 No Solution Yet Submitted by Bractals Rating: 5.0000 (2 votes)

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 Possible solution Comment 1 of 1
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, AiSi, (i = 1,2,3) are

concurrent.

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

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

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