In triangle AXZ, angle XAZ measures 80, segment XY bisects angle AXZ, and ZY bisects AZX. What is the measure of angle XYZ?
The answer should be 130 but I don't see how to obtain it.
From the given, we're only given the one angle of 80 at angle A and the two bisectors (I, too, thought about using the properties of isosceles triangles but it seems there is not enough info to do so which is why I'm stuck).
You know (angle XAZ=80°) and the triangle is isosceles, so that means (angle AXZ) =50° =(angle AZX). Because XY bisects angle AXZ that means the bisector divides the (angle AXZ = 50°) in two equal angles of 25°. So if you make the addition: 180°-(25°+25°)=130°=(angle XYZ).
The calculations certainly make sense, but how can you deduce that either triangle would be isosceles? That is the segments that are given that bisect the base angles of triangle AXZ are necessarily congruent base angles..
Hello, dannyc!
$\displaystyle \text{In }\Delta AXZ,\;\angle XAZ\,=\,80^o.$
$\displaystyle \text{Segment }XY\text{ bisects }\angle AXZ\text{, and }ZY\text{ bisects }\angle AZX.$
$\displaystyle \text{What is the measure of }\angle XYZ\,?$
Code:A * * * *80d* * * * Y * * * * * * @ * * * * * * * * * * ** 1 2** X * * * * * * * Z
Let: .$\displaystyle \begin{Bmatrix}\angle X \:=\:\angle AXZ \\ \angle Z \:=\:\angle AZX \end{Bmatrix}\qquad \begin{Bmatrix}\angle 1 &=^& \angle YXZ &=& \frac{1}{2}\angle X\\ \angle 2 &=& \angle YZX &=& \frac{1}{2}\angle Z\\ \theta &=& \angle XYZ \end{Bmatrix}$
$\displaystyle \text{In }\Delta AXZ\!:\;\angle X + \angle Z + 80^o \:=\:180^o $
. . . . . . . . . . . . . $\displaystyle \angle X + \angle Z \:=\:100^o$
$\displaystyle \text{Divide by 2:}$ . . $\displaystyle \tfrac{1}{2}\angle X + \tfrac{1}{2}\angle Z \:=\:50^o$ .[1]
$\displaystyle \text{In }\Delta XYZ\!:\;\angle 1 + \angle 2 + \theta \:=\:180^o$
. . . . . .$\displaystyle \tfrac{1}{2}\angle X + \tfrac{1}{2}\angle Z + \theta \:=\:180^o$
Substitute [1]: . $\displaystyle 50^o + \theta \:=\:180^o$
. . . Therefore: . . = . . $\displaystyle \theta \:=\:130^o$
Thought of just to add on a point as well.
The point of intersection y is called in-centre. If you draw a circle taking it as the centre and perpendicular distance from it to any of the side the circle will pass through the other two points of the triangle