Advanced High-School Mathematics

SECTION 1.2 Triangle Geometry 17

3. Given 4 ABC and assume thatX is on (BC),Y is on (AC) and
   Z is on (AB). Assume that the Cevians (AX) (BY),and (CZ)
   are concurrent, meeting at the pointP. Show that
   PX
   AX

+

PY

BY

+

PZ

CZ

= 1.

4. Given the triangle 4 ABCwith incenterP, prove that there exists
   a circleC (called theincircleof 4 ABC) with centerP which is
   inscribed in the triangle 4 ABC. The radius r of the incircle is
   often called theinradiusof 4 ABC.


5. Let 4 ABC have side lengthsa = BC, b = AC, and c = AB,
   and letr be the inradius. Show that the area of 4 ABC is equal
   to r(a+ 2 b+c). (Hint: the incenter partitions the triangle into three
   smaller triangles; compute the areas of each of these.)


6. Given the triangle 4 ABC. Show that the bisector of theinternal
   angle bisector atAand the bisectors of theexternalangles atB
   andC are concurrent.


7. Given 4 ABCand pointsX, Y,and
   Zin the plane such that

∠ABZ = ∠CBX,

∠BCX = ∠ACY,

∠BAZ = ∠CAY.

Show that (AX), (BY), and (CZ)

are concurrent.

Z

Y

X

C

B

A

8. There is another notion of “center” of the triangle 4 ABC. Namely,
   construct the linesl 1 , l 2 ,andl 3 so as to be perpendicular bisectors
   of [AB],[BC], and [CA], respectively. After noting that Ceva’s
   theorem doesn’t apply to this situation, prove directly that the
   lines l 1 , l 2 , and l 3 are concurrent. The point of concurrency is
   called thecircumcenterof 4 ABC. (Hint: argue that the point
   of concurrency of two of the perpendicular bisectors is equidistant
   to all three of the vertices.) If P is the circumcenter, then the
   common value AP = BP = CP is called the circumradius