Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Monday, May 4, 2015

### Geometry Problem 1118: Right Triangle, Angle Trisection, Concyclic Points, Cyclic Quadrilateral

Labels:
concyclic,
cyclic quadrilateral,
right triangle,
trisection

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http://s12.postimg.org/a4sui5wi5/pro_1118.png

ReplyDeleteSince B3C4 and CC4 bisect angles B1B3C and B3CB1 ( see problem 1116)

So B1C4 will bisect angle B3B1C

And ∠ (B3B1C4)= ½ . 60= 30 degrees

And ∠ (C3B1C4)= ∠ ( C3B3C4)= 90 => B3, C3, B1, C4 are concyclic

Similarly we also have B1A4 bisect angle AB1B3 and points A3, B3, A4 ,B1 are concyclic

Kindly refer my proofs for Problems 1114 thro to 1117. We have seen that < C3B1A4 = A4B1B2=< B2B1C4 = C4B1A3 = 30

ReplyDeleteHence < C3B1C4 is 90 and since < C3B3C4 = 90 C3B3C4B1 is cyclic

Similarly A4B3A3B1 is cyclic

Sumith Peiris

Moratuwa

Sri Lanka

Another problem arising from this-

ReplyDeleteProve B1B2A4C3 is cyclic and so is B1A3C4B2

Sumith Peiris

Moratuwa

Sri Lanka