Question

name two pairs of congruent angles in the figure. justify the answer.
select all that apply
a. ∠fph and ∠epg according to the corresponding angles theorem
b. ∠epg and ∠fph according to the equal complements theorem
c. ∠eph and ∠fph according to the equal supplements theorem
d. ∠fpg and ∠eph according to the right angles congruent theorem
e. ∠fpg and ∠eph according to the corresponding angles theorem
f. ∠epg and ∠fpg as given in the figure
g. ∠eph and ∠epg according to the linear pair theorem
h. ∠fph and ∠fpg according to the right angles congruent theorem

Explanation:

Answer:

F. \( \angle EPG \) and \( \angle FPG \) as given in the figure
H. \( \angle FPH \) and \( \angle FPG \) according to the Right Angles Congruent Theorem

(Note: Assuming the figure shows \( \angle EPG \cong \angle FPG \) (given) and \( \angle FPH \) and \( \angle FPG \) are right angles (so congruent by Right Angles Congruent Theorem). Also, checking the options, F and H are valid. Another possible valid pair could be B if \( \angle EPG \) and \( \angle FPH \) are complements of the same angle, but based on typical diagram interpretations, F (given congruent angles) and H (right angles congruent) are clear. Also, D: \( \angle FPG \) and \( \angle EPH \) – if \( \angle EPH \) is a right angle (since H-P-G is a straight line, and if \( \angle EPG \) and \( \angle FPG \) are equal, maybe \( \angle EPH \) and \( \angle FPH \) are right angles? Wait, maybe rechecking:

Wait, H-P-G is a straight line, so \( \angle HPG = 180^\circ \). If \( \angle EPG = \angle FPG \) (given, option F), then \( \angle EPG + \angle FPG = \angle EPF \), but maybe \( \angle EPH \) and \( \angle FPH \) are right angles? Wait, no, \( \angle FPH \): H-P-F – if H-P-G is straight, and P-F is a ray, maybe \( \angle FPH \) is a right angle? Wait, maybe the correct options are F (given \( \angle EPG \cong \angle FPG \)) and H (\( \angle FPH \) and \( \angle FPG \) are right angles, so congruent by Right Angles Congruent Theorem) or F and B? Wait, maybe the intended answers are F (given congruent angles) and H (right angles congruent) or F and B. But based on the options, F is "as given in the figure" (so if the figure marks \( \angle EPG \) and \( \angle FPG \) as equal, that's valid). H: \( \angle FPH \) and \( \angle FPG \) – if both are right angles (since H-P-G is straight, and if \( \angle EPG = \angle FPG \), maybe \( \angle EPH = \angle FPH = 90^\circ \)? Wait, maybe the correct options are F and H, or F and B. But let's go with the most straightforward: F (given congruent angles) and H (right angles congruent) or F and B. Wait, the Equal Complements Theorem: if two angles are complements of the same angle, they are congruent. If \( \angle EPG \) and \( \angle FPH \) are both complements of \( \angle EPH \) or something, but maybe the intended answers are F and H. So the final answer is F and H (and possibly B, but let's check the options again. The problem says "name two pairs", so two options. So F and H, or F and B. But based on the options, F is "as given", H is right angles congruent. So:

F. \( \angle EPG \) and \( \angle FPG \) as given in the figure
H. \( \angle FPH \) and \( \angle FPG \) according to the Right Angles Congruent Theorem

(Alternatively, B: \( \angle EPG \) and \( \angle FPH \) according to Equal Complements Theorem, if they are complements of the same angle. But without the figure, assuming the typical diagram where H-P-G is straight, P-E and P-F form angles with \( \angle EPG = \angle FPG \) (given), and \( \angle FPH \) and \( \angle FPG \) are right angles (so H is a right angle). So F and H are valid.)