Advanced High-School Mathematics

18 CHAPTER 1 Advanced Euclidean Geometry

of the triangle 4 ABC. (This is because the circumscribed circle

containingA, B, andC will have radiusAP.)

9. 4 ABC has side lengths AB = 21, AC = 22, and BC = 20.
   Points D and E are on sides [AB] and [AC], respectively such
   that [DE]‖[BC] and [DE] passes through the incenter of 4 ABC.
   ComputeDE.


10. Here’s another proof of Ceva’s the-
    orem. You are given 4 ABC and
    concurrent Cevians [AX],[BY],
    and [CZ], meeting at the pointP.
    Construct the line segments [AN]
    and [CM], both parallel to the Ce-
    vian [BY]. Use similar triangles to
    conclude that
    AY
    Y C

=

AN

CM

,

CX

XB

=

CM

BP

,

BZ

ZA

=

BP

AN

,

N M

P

Y

Z X

C

B

A

and hence that

AZ

ZB

·

BX

XC

·

CY

Y A

= 1.

11. Through the vertices of the triangle 4 PQRlines are drawn lines
    which are parallel to the opposite sides of the triangle. Call the
    new triangle 4 ABC. Prove that these two triangles have the same
    centroid.


12. Given the triangle 4 ABC, let C be the inscribed circle, as in
    Exercise 4, above. Let X, Y, and Z be the points of tangency
    of C(on the sides [BC],[AC],[AB], respectively) and show that
    the lines (AX),(BY), and (CZ) are concurrent. The point of
    concurrency is called theGergonne pointof the circleC. (This
    is very easy once you note thatAZ=Y Z, etc.!)


13. In the figure to the right, the dotted
    segments represent angle bisectors.
    Show that the points P, R, andQ
    are colinear.