Question

Altitude line segment N P. is drawn in triangle MNO. Which answer choice can prove that line segment N P. is the geometric mean of line segment M P. and line segment P O.?

M,. N,. O,. P,.
A
right angles M P N, N P O, and O N M are congruent. Triangles N M O and N M P share angle N M P, and triangles N M O and N O P share angle N O P. So, triangles N M O, N M P, and N O P are similar by the A, A Similarity Theorem. So, M P over N P equals N P over O P,.
B
since right angles are congruent, angle M P N, angle N P O, and angle O N M are congruent. Segment N P bisects angle M N O, so angle M N O is congruent to angle O N P. So triangles N P M and O N P are similar, and M P over N P equals N P over O P,.
C
segment N P bisects angle M N O, so angle M N P equals 1 over 2 angle M N O and angle O N P equals 1 over 2 angle O N M. By substitution, M P over N P equals N P over O P.
D
By the Alternate Interior angle theorem, angle M N P equals 1 over 2 angle N P O and angle O N P equals 1 over 2 angle N P M. By substitution, M P over N P equals N P over O P.

Answer 1

A.

Reason: Let NP be the altitude from N to MO, so NP ⟂ MO. Then ∠MPN and ∠NPO are right angles and therefore congruent. Also
- ∠NMP = ∠NMO (MP lies on MO), so triangle MNP and triangle MNO share that acute angle,
- ∠NOP = ∠NOM, so triangle NOP and triangle MNO share that acute angle.

Thus by AA, triangles MNP, NOP, and MNO are similar. From the similarity of the two small right triangles,
MP / NP = NP / PO,
so NP^2 = MP · PO, i.e. NP is the geometric mean of MP and PO.