## 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.

```

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