## Side of a Parallelogram

In parallelogram ABCD, the bisector of angle ABC intersects AD
at P. If PD=5, BP=6, and CP=6, what is the value of AB?

Source:
The Geometry Forum , July 1, 1996.

Ken's Solution: AB = 4
Let angle ABC = 2x. Then ABP = PBC = x. And PAB = 180-2x.

In triangle PAB, Angle APB = 180 - PAB - ABP = x. So triangle PAB
is isosceles, with base BP=6.

Since BP=PC=6, triangle BPC is also isosceles with base BC, and sides
of length 6. So angle BCP = x, and angle BPC = 180-2x.

By similar angles, triangles ABP and BPC are similar. The following
ratio leads to the solution (Note that AB=AP, and AD=BC):

AB 6
----- = ----- ==> (AB)^2 + 5(AB) - 36 = 0
6 AB+5

Using the quadratic formula for AB, we find AB = 4.

Solution recieved from Bill Chapp:
Refer to the following, badly drawn, ascii diagram for reference:
B --------------- C
\ ` '\
\ 6` '6 \
\ ` ' 5 \
A ------.-------- D
P
Given: BP bisects angle ABC
PD=5, BP=6, and CP=6.
Find AB
1) Since BP bisects angle ABC, angle ABP = angle PBC.
2) Since BP = CP, anlge PBC = angle BCP.
3) By opposite inner angles, angle CPD = angle BCP.
3a) Let Theta = ABP = PBC = BCP = CPD
4) Sum of angles in triangle ABP: ABP + BPA + PAB = 180.
4a) Theta + BPA + PAB = 180
5) Sum of angles on line APD: BPA + BPC + CPD = 180.
5a) Theta + BPA + BPC = 180
6) Eqn 4a) minus Eqn 5a) ==> PAB - BPC = 0
6a) PAB = BPC
7) Since angle PAB = angle BPC and angle ABP = angle PBC,
angle PBA = angle BCP, therefore triangle ABP is similar
to triangle PBC.
7a) AB/BP = BP/BC
7b) Triangle ABP is isosceles w/ AB = AP
8) AD = AP + PD, and AD = BC, therefore BC = AB + PD
9) Substituting 8) and given lengths into 7a),
AB/6 = 6/(AB+5)
AB*(AB + 5) = 36
AB^2 + 5*AB - 36 = 0
(AB + 9)*(AB - 4) = 0
AB = -9 or 4
Choosing positive value gives AB = 4.

Mail to Ken