Question

for # 13 - 14, using the given information, determine if the triangles are congruent. if so, name the triangles and state the reason.

1. given: z is the midpoint of \\( \overline { w y } \\)
2. given: \\( \overline { a d } \\) bisects \\( \angle b a c \\)

for # 15 - 18, solve the algebra problems below.

1. solve for x, y and z.
2. given \\( \triangle f h s \cong \triangle g e o \\), solve for x and y.

Explanation:

Step1: Find \(x\) in problem 15

Since the triangle with \(x\) and \(36^{\circ}\) is isosceles (two sides are equal), \(x = 36^{\circ}\).

Step2: Find \(y\) in problem 15

Using the exterior - angle theorem for the large triangle. The exterior angle \(y\) is equal to \(46^{\circ}+36^{\circ}\). So \(y=82^{\circ}\).

Step3: Find \(z\) in problem 15

Using the angle - sum property of a triangle (\(180^{\circ}\) in a triangle). For the triangle with \(z\), \(y\), and \(z\): \(z = 180^{\circ}-36^{\circ}-82^{\circ}=62^{\circ}\).

Step4: Use congruent - triangle properties for problem 16

Since \(\triangle FHS\cong\triangle GEO\), corresponding angles and sides are equal.
\(\angle F=\angle G = 22^{\circ}\), \(\angle S=\angle O = 94^{\circ}\), and \(FS = GO\), \(HS=EO\).
For the sides:
\(- 2x + 12y=94\) and \(4x + 6y=22\).
Multiply the second equation \(4x + 6y=22\) by \(2\) to get \(8x+12y = 44\).
Subtract \(-2x + 12y=94\) from \(8x + 12y=44\):
\((8x+12y)-(-2x + 12y)=44 - 94\)
\(8x+12y + 2x-12y=-50\)
\(10x=-50\), so \(x=-5\).
Substitute \(x = - 5\) into \(4x+6y = 22\):
\(4\times(-5)+6y=22\)
\(-20 + 6y=22\)
\(6y=42\), so \(y = 7\).

Answer:

13)
Congruent? YES
\(\triangle WXZ\cong\triangle YXZ\)
By: SAS
14)
Congruent? YES
\(\triangle BAD\cong\triangle CAD\)
By: AAS
15)
\(x = 36\)
\(y = 82\)
\(z = 62\)
16)
\(x=-5\)
\(y = 7\)