Question

for items 8–10, use $\triangle def$ shown.

1. what is $m\angle def$? $\boldsymbol{50}^\circ$
2. select all the descriptions for $\overline{ge}$.

a. angle bisector
b. median
c. perpendicular bisector
d. altitude
e. hypotenuse

1. select all the points that segment $\overline{ge}$ contains.

a. circumcenter
b. incenter
c. orthocenter
d. centroid
e. midpoint of edge $de$

Explanation:

Item 8 (already solved as 50°, but let's confirm the process)

Step1: Identify triangle properties

Since \( DE = EF \) (marked with congruency ticks), \( \triangle DEF \) is isosceles with \( \angle DEG=\angle FEG \). So \( 3y + 4 = 5y - 10 \).

Step2: Solve for \( y \)

\( 3y + 4 = 5y - 10 \)
Subtract \( 3y \): \( 4 = 2y - 10 \)
Add 10: \( 14 = 2y \)
Divide by 2: \( y = 7 \).

Step3: Find \( \angle DEF \)

\( \angle DEF=(3y + 4)+(5y - 10)=8y - 6 \). Substitute \( y = 7 \): \( 8(7)-6 = 56 - 6 = 50^\circ \).

• A. Angle bisector: \( \angle DEG=\angle FEG \), so \( GE \) bisects \( \angle DEF \).
• B. Median: \( G \) is midpoint of \( DF \) (marked with ticks), so \( GE \) is a median.
• C. Perpendicular bisector: In isosceles triangle, median to base is perpendicular bisector. \( GE \perp DF \) (since isosceles, altitude/median/angle bisector coincide), so it’s a perpendicular bisector of \( DF \).
• D. Altitude: In isosceles triangle, median to base is altitude ( \( GE \perp DF \) ).
• E. Hypotenuse: Hypotenuse is in right triangles, \( \triangle DEF \) isn’t necessarily right, and \( GE \) is a segment from vertex to base, not a hypotenuse.

So A, B, C, D apply.

• A. Circumcenter: In a triangle, circumcenter is intersection of perpendicular bisectors. \( GE \) is perpendicular bisector of \( DF \), so circumcenter lies on \( GE \).
• B. Incenter: Incenter is intersection of angle bisectors. \( GE \) is angle bisector of \( \angle DEF \), so incenter lies on \( GE \).
• C. Orthocenter: Orthocenter is intersection of altitudes. \( GE \) is an altitude, so orthocenter lies on \( GE \).
• D. Centroid: Centroid is intersection of medians. \( GE \) is a median, so centroid lies on \( GE \).
• E. Midpoint of \( DE \): \( GE \) connects \( E \) to midpoint of \( DF \), not midpoint of \( DE \), so no.

Thus A, B, C, D apply.

Answer:

\( 50^\circ \)

Item 9