Question

HELP ME GENIUSES!!!!!

Given: AC is parallel to BD, and AB is parallel to CD.
AC is perpendicular to CD
Prove: angle PCQ is complementary to angle ABC.

Answer 1

Proof: Since AC is perpendicular to CD, angle mOCQ = 90° by the definition of perpendicular lines. By angle addition, we can say angle mOCQ = angle mOCP + angle mPCQ. But since angle mOCQ = 90°, angle mOCP + angle mPCQ = 90° by the Transitive Property of Equality.

[Missing Step]

By the definition of congruent angles, angle mOCP = angle mABC. This leads to
angle mABC + angle mPCQ = 90° by the Transitive Property of Equality. So, based on the definition of complementary angles, angle PCQ is complementary to angle ABC.

Answer 2

What is the missing step in the given proof?
A. Angle PCQ and angle ACP are supplementary by the Linear Pair Theorem.
B. For parallel lines cut by a transversal, corresponding angles are congruent, so angle ACB is congruent angle PCQ.
C. Angle OCP is congruent to angle BCD by the Vertical Angles Theorem.
D. For parallel lines cut by a transversal, corresponding angles are congruent, so angle OCP is congruent to angle ABC.
E. For parallel lines cut by a transversal, corresponding angles are congruent, so angle OCA is congruent to angle CBD.