Triangle ABC has points P and Q on BC in the order B,P,Q,C. Lines AP and AQ trisect angle BAC.
Part 1: Given BP=3, PQ=4, and QC=6, determine the lengths of AB and AC.
Part 2: More generally, express AB and AC in terms of BP, PQ, and QC.
Part 2:
Let b = AB, c = AC, p = AP, q = AQ,
x = BP, y = PQ, z = QC
AP bisects angle BAQ and AQ bisects angle PAC.
Therefore,
p = c*y/z (1)
and
q = b*y/x (2)
Applying the cosine law to triangles BAP, PAQ,
and QAC gives
x^2 = b^2 + p^2  2*b*p*cos(angle BAP)
and
y^2 = p^2 + q^2  2*p*q*cos(angle PAQ)
and
z^2 = q^2 + c^2  2*q*c*cos(angle QAC)
Angle BAP = angle PAQ. Therefore,
b^2 + p^2  x^2 p^2 + q^2  y^2
 =  (3)
b q
Angle QAC = angle PAQ. Therefore,
c^2 + q^2  z^2 p^2 + q^2  y^2
 =  (4)
c p
Substituting p and q from (1) and (2) into (3)
gives
b^2*z^2  c^2*x*y = x^2*z^2 (5)
Substituting p and q from (1) and (2) into (4)
gives
b^2*y*z + c^2*x^2 = x^2*z^2 (6)
Solving (5) and (6) for b^2 and c^2
b^2 = x^2*z*(x + y)/(x*z  y^2)
c^2 = x*z^2*(y + z)/(x*z  y^2)
or
AB = BP*sqrt(BQQC/UV^2)
AC = QC*sqrt(BPPC/UV^2)
where UV^2 = BPQC  PQ^2
QED for Part 2
Part 1:
BP = 3, PQ = 4, and QC = 6
BQ = BP + PQ = 3 + 4 = 7
PC = PQ + QC = 4 + 6 = 10
UV^2 = 3*6  4^2 = 2
AB = 3*sqrt(7*6/2) = 3*sqrt(21)
~= 13.7477
AC = 6*sqrt(3*10/2) = 6*sqrt(15)
~= 23.2379
QED for Part 1

Posted by Bractals
on 20120422 17:57:41 