Question 147100
In Triangle ABC, it is given that CA=CB. Points P and Q are marked on segments CA and CB respectively, so that angles CBP and CAQ are the same size. Prove that CP=CQ.
<pre><font size = 4 color = "indigo"><b>
{{{drawing(200,300,-3, 3,-1,6,
locate(-2.3,0, A), locate(2,0,B), locate(0,5.4,C),
triangle(-2,0,2,0,0,5),

triangle(-2,0,-1,5/2,2,0), locate(-1.4,2.7,P), locate(1.2,2.7,Q),

triangle(2,0,1,5/2,-2,0) )}}}

<font face = "symbol">Ð</font>CBP = <font face = "symbol">Ð</font>CAQ      Given

  CA = CB        Given 

  <font face = "symbol">Ð</font>C = <font face = "symbol">Ð</font>C        Identity

<font face = "symbol">D</font>CBP <font face = "symbol">@</font> <font face = "symbol">D</font>CAQ      Angle-Side-Angle

  CP = CQ        Corresponding parts of congruent triangles

Edwin</pre>