294 CHAPTER (^8) GEOMETRY FOR AMERICANS
cut AB in D. Prove that the length of CD is
2ab cos (C /2)
a+b
8.4. 10 (Canada 1986) A chord ST of constant length
slides around a semicircle with diameter AB. M is the
midpoint of ST and P is the foot of the perpendicular
from S to AB. Prove that angle SPM is constant for all
positions of ST.
8.4. 11 Finish the proof of Ceva's theorem by estab
lishing the converse: if the product of the ratios (see
picture on page 288) is
AF BD CE
FB DC EA
= I,
then the cevians AD, BE, C F are concurrent. Hint:
proof by contradiction.
8.4. 12 (Putnam 1997) A rectangle, HOMF, has sides
HO = 11 and OM = 5. A triangle ABC has H as the
intersection of the altitudes, 0 the center of the cir
cumscribed circle, M the midpoint of BC, and F the
foot of the altitude from A. What is the length of BC?
8.4. 13 Use Ceva's theorem to prove the following al
ready proven concurrences.
(a) The three angle bisectors of a triangle all meet
in a single point.
(b) The three medians of a triangle all meet in a sin
gle point.
(c) The three altitudes of a triangle all meet in a
single point.
8.4. 14 (AIME 2003 ) Find the area of rhombus ABCD
given that the radii of the circles circumscribed around
triangles ABD and ACD are 12.5 and 25, respectively.
8.4. 15 (Canada 1990) Let ABCD be a convex quadri
lateral inscribed in a circle, and let diagonals AC and
BD meet at X. The perpendiculars from X meet the
sides AB, BC, CD, DA at A', 8 ', C', D', respectively.
Prove that
A'B' +C'D' =A'D' +B'C'.
8.4. 16 (Bay Area Mathematical Olympiad 200 I) Let
J H IZ be a rectangle, and let A and C be points on sides
Zl and ZJ, respectively. The perpendicular from A to
CH intersects line HI in X, and the perpendicular from
C to AH intersects line H J in Y. Prove that X, Y and Z
are collinear.
8.4. 17 The Orthic Triangle. Let ABC be a triangle.
The triangle whose vertices are the feet of the three
altitudes of ABC is called the orthic triangle of ABC.
(a) Show that the measures of the angles of the or
thic triangle of ABC are 11: -2A, 11: -2B, 11: -2C.
(b) Show that if ABC is acute (and hence its orthic
triangle lies inside it), the altitudes of ABC bi
sect the angles of the orthic triangle. In other
words, the orthocenter of an acute triangle is the
incenter of its orthic triangle.
8.4. 18 Draw lines from each vertex of a parallelogram
to the midpoints of the two opposite sides. An eight
sided figure will appear in the center of the parallelo
gram. Find the ratio of its area to that of the parallelo
gram.
8.4. 19 Let ABCD be a convex quadrilateral ("convex"
means that every line segment joining two vertices lies
entirely on or inside the quadrilateral; a non-convex
polygon has "indentations"). Let P be an arbitrary
point on side AB. Draw a line through A that is par
allel to PD. Likewise, draw a line through B parallel
to Pc. Let Q be the intersection of these two lines.
Prove that
[DQCJ = [ABCD].
8.4.20 Consider the diagram used in the proof of
Ceva's theorem on page (^288). Prove that
8.4.21 The Method of "Weights." In Example 8.4.2 on
page 286, comparing areas gave us information about
segment ratios. Here is another method, using infor
mal physical intuition. Imagine placing weights on
the vertices of triangle ABC. (Refer to the figure on
page 286). Start with a weight of 2 units at vertex B.
We can place any weight we want at A, but if we as
sume that F is the balancing point, we must place 4
units at A, since the product of each weight times the
distance to the balance point must be equal. We can
then imagine that the two weights at A and B are equiv
alent to a single weight of 6 units, at F.
Continue this process, now looking at side CB,
with a balancing point at D. We already have a weight
of 2 units at B, so this forces a weight of I at C, in or
der to balance. We place the sum, 3, at the balancing
point. Here is the result.
ann
(Ann)
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