practice problems pending Q2 Q3

 Q1. The internal angle bisector of ∠A of △ABC meets BC at P and (b = 2c) in the usual notation. Prove that ((9AP^2 + 2a^2)) is an integral multiple of (c^2).

S1.

Let PC = n, BP = m, AP = d
We need to prove:
9d^2 + 2a^2 = k.c^2
where 'k' is an integer.

c/m = b/n by Angle Bisector theorem.
=> 2m = n
=> 3m = a
=> m = a/3, n = 2a/3
Using Stewart's theorem:

man + dad = bmb + cnc = m.4c^2 + nc^2 = a(mn + d^2) = c^2 (4m + n)
=> a(n^2/2 + d^2) = c^2(3n)
=> a(4a^2/18 + d^2) = c^2.2a/3
=> 4c^2/3 = (2a^2/9 + d^2)
=> 12c^2 = (2a^2 + 9d^2)
H.P.

Q2. In (\triangle ABC), in the usual notation, the area is bc/2 sq. units. (AD) is the median to (BC).

Prove that

Angle ABC = (Angle ADC)/2
S2.
Area = 1/2.bc.sin(A) => A = 90 deg.
So D is the circumcenter.
Chord AC subtends angle ADC at center and ABC at circumference.
H.P.

Q3.

Let C1 be any point on (AB) of triangle ABC.

Draw CC1 meeting AB at C1. The lines through A and B parallel to CC1 meet BC produced and AC produced at A1 and B1, respectively.

Pr. th.
1/CC1 = 1/AA1 + 1/BB1
S3.

AC1C ~ ABB1__[1]
BC1C ~ BAA1___[2]

From [1]:
AC1/AB = C1C/BB1 = AC/AB1

From [2]:
BC1/BA = C1C/AA1 = BC/BA1

Add them:
(AC1 + C1B)/AB = C1C(1/AA1 + 1/BB1) = 1
H.P.