# Thread: Triangle Question

1. ## Triangle Question

In a triangle ABC where <A=120 degrees, Q and P are points on BC such that BP=BA and CQ=CA, where Q lies between B and P. Prove that <PAQ=30 degrees.

I really could use a lot of help on this one.

2. Hello, TheNcredibleHulk!

Did you make a sketch?

In triangle $\displaystyle ABC$ where $\displaystyle \angle A=120^o,\;P$ and $\displaystyle Q$ are points on $\displaystyle BC$
such that: $\displaystyle BP=BA$ and $\displaystyle CQ=CA$, where $\displaystyle Q$ lies between $\displaystyle B$ and $\displaystyle P.$
Prove that: .$\displaystyle \angle PAQ\,=\,30^o$
Code:
    B o
* *
*   *
*  θ  *
*       *    Q
*         o    P
*           o
*       *     *
*   *           *
o  *  *  *  *  *  o C
A

Draw line segment $\displaystyle AQ$.

Let $\displaystyle \angle B \,=\,\theta$ . . . Then $\displaystyle \angle C \,=\,60-\theta$

In isosceles $\displaystyle \Delta ABP\!:\;\angle BPA = \angle BAP$

. . $\displaystyle \theta + \angle BPA + \angle BAP \:=\:180^o \quad\Rightarrow\quad \angle BPA \:=\:90 - \tfrac{1}{2}\theta \:=\:\angle QPA$

In isosceles $\displaystyle \Delta CAQ\!:\;\angle CQA = \angle CAQ$

. . $\displaystyle (60-\theta) + \angle CQA + \angle CAQ \:=\:180^o \quad\Rightarrow\quad \angle CQA \:=\:60 + \tfrac{1}{2}\theta \:=\:\angle PQA$

In $\displaystyle \Delta APQ\!:\;\angle PAQ + \angle PQA + \angle QPA \:=\:180^o$

Therefore: .$\displaystyle \angle PAQ + \left(90^o-\tfrac{1}{2}\theta\right) + \left(60^o +\tfrac{1}{2}\theta\right) \:=\:180^o \quad\Rightarrow\quad \angle PAQ \:=\:30^o$