Butterfly problem ?

[lvl107]

Hello, Everyone.
From the following :
Given :
Let ABC be a triangle with incircle ( I ), and let ( I ) touch BC, CA, AB at A', B', C', respectively.
Drop perpendicular from A' onto B'C' at A''. Lines BA'' and CA meet at B''. Lines CA'' and AB meet
at C''.
Prove :
C'C'' + B'B'' = C''B''

( dark-green + green = red )

    Thanks in advance for your time and help.
        Best Regards.

             Loi.

[idahoan]

Please attach a MathCAD worksheet that shows the problem you are experiencing.  Then perhaps we can assist.

[Werner_E]

Actually it does not matter where you choose the point A'' on the line B'C' ! 

B''C'' will always be a tangent to the circle and so the two green lines will add up to the red one.

[lvl107]

   Many thanks for your time and hints, idahoan and Werner. @Werner_E , your mark, the above, is a more general statement !

And the butterfly problem should be stated as the following:

     Let ABC be a triangle with incircle ( I ), and let ( I ) touch AC, AB at B', C', respectively. M is a
point on segment B'C'. Lines BM and CA meet at B''. Lines CM and AB meet at C''.
Prove :
C'C'' + B'B'' = C''B''.  

( dark-green + green = red )

    And I still got stuck proving B''C'' be a tangent to the incircle ( I ).

        Best Regards.

[ttokoro]

Make △OB"B'≡△OB"P and make △OC"C'≡△OC"P.

Length of OB'=OC'=OP=radius r of the circle.

Angle of ∠OB'B"=∠OPB"=∠OC'C"=∠OPC"=90 deg.

Therefore, P of △OB"P and △OC"P is same point and on the circle.

And B''C'' be a tangent to the incircle ( I ).  ....Still not solve. 

Make △B'C'P to circumscribed circle. Point P is also on the incircle (I).

This center of circumscribed circle, O must show above properties.   ....Still not solve.

[lvl107]

   Many thanks for your respond, Ttokoro. :

   Best Regards.

[ttokoro · Accepted solution]

For my above picture, at first point P on the circle (1) is selected between circular arc c'Pb'.

Then make tangent line c"Pb" at point P on the circle. This is the red line.

Then 

Make △OB"B'≡△OB"P and make △OC"C'≡△OC"P.

Length of OB'=OC'=OP=radius r of the circle.

Angle of ∠OB'B"=∠OPB"=∠OC'C"=∠OPC"=90 deg. (Because the tangent line must be perpendicular to radius r.)

So

If we can show the intersection point M of line cc" and bb" is also on the line c'b', it is solved.

[ttokoro]

If we can show the intersection point M of lines cc" and bb" is also on the line c'b', and it is also the intersection point of lines A'P and b'c'.
Then, it is solved.
Using Mathcad numerical calculation with using your 2020's puzzle, it is true.

[ttokoro]

 About 12 digits are same but not equal.

[ttokoro]

Attach Prime 8 worksheet to show all.

Tokoro. 

[lvl107]

Many, many thanks for the extra response, @ttokoro . And with following you hints, I have just done a very small thing, but with all of calculation was done with only by symbolic ( NOT NUMERIC ) then :

   Best Regards.

       Loi.

[lvl107]

  Ttokoro, To your extra response, one very small thing, I have just done now : M is on the extension line of B'C'.

 

   Best Regards.

       Loi.

[ttokoro]

Thanks. I also find it. Chage my T value of combo box, such as 1.2, Prime 8 also shows it.

Tokoro.   

[lvl107]

Ttokoro, in your case #2 then :

Best Regards.

     Loi.