Geometry with Trigonometry

Sec. 10.11 A case of Pascal’s theorem, 1640 181

thenP,Q,R,Sare the vertices of a parallelogram.

10.2 IfZ 1 ∼z 1 ,Z 2 ∼z 2 andZ 3 ∼z 3 are non-collinear points show that

z 1 (z ̄ 3 −z ̄ 2 )+z 2 (z ̄ 1 −z ̄ 3 )+z 3 (z ̄ 2 −z ̄ 1 )= 4 iδF(Z 1 ,Z 2 ,Z 3 )= 0.

10.3 LetA∼a,B∼b,C∼cbe non-collinear points andP∼pa point such that

AP,BP,CPmeetBC,CA,ABatD∼d,E∼e,F∼f, respectively. Show for

sensed ratio that

BD

DC

=

p(b ̄−a ̄)+p ̄(a−b)+ba ̄−ab ̄

p(a ̄−c ̄)+p ̄(c−a)+ac ̄−ca ̄

,

and hence prove Ceva’s theorem that

BD

DC

CE

EA

AF

FB

= 1.

10.4 LetA∼a,B∼b,C∼cbe non-collinear points. Given any pointP∼p,show

that as(c−a)/(b−a)is non-real there exist unique real numbersyandzsuch

thatp−a=y(b−a)+z(c−a),andsop=xa+yb+zcwherex+y+z=1.

Show that ifAPmeetsBCit is in a pointD∼dsuch that

d=

1

1 +r

b+

r

1 +r

c,

wherer=z/y. Hence prove Ceva’s theorem that ifD∈BC,E∈CA,F∈AB

are such thatAD,BE,CFare concurrent, then

BD

DC

CE

EA

AF

FB

= 1.

10.5 LetA∼a,B∼b,C∼cbe non-collinear points. IfD∼d,E∼ewhere

d=

1

1 +λ

b+

λ

1 +λ

c,e=

1

1 +μ

c+

μ

1 +μ

a,

andDEmeetsABit is in a pointF∼fwhere

f=

1

1 +ν

a+

ν

1 +ν

b,

andλμν=−1.

Prove Menelaus’ theorem that ifD∈BC,E∈CA,F∈ABare collinear, then

BD

DC

CE

EA

AF

FB

=− 1.