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

Draw line segment $AQ$.

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

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

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

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

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

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

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