Prove the Following:

Question

Given line PC is perpendicular to line KM, with K-P-M. A and B are points on the same side of line KM as C, but A and B are on opposite sides of line PC. A is on the same side of line PC as K. Triangle ACP is congruent to triangle BCP. Prove that angle KPA is congruent to angle MPB.

Answer 1

Extends line segment CA so that is passes through K;

Extends line segment CB so that is passes through M;

Triangle ACP is congruent to triangle BCP is GIVEN;

then angle APC = angle BPC by CPCTC...

PC is perpendicular to KM is GIVEN

angle CPK and angle CPM are right angles by definition of perpendicular

angle CPK = 90 and angle CPM = 90 by definition of right angles

angle CPK = angle CPM by substitution/transitive

angle CPK = angle APC + APK and angle CPM = angle BPC + angle BPM by angle addition postulate

angle APC + APK = angle BPC + angle BPM

angle APK = angle BPM by subtraction property of equality

angle CAP and angle KAP are supplementary linear pairs,

as are angles CBP and BPM;

angle CAP + angle KAP = 180 AND

angle CBP + angle BPM = 180 by definition of supplementary

angle CAP + angle KAP = angle CBP + angle BPM by transitive/substitution

angle CAP = angle CBP by CPCTC

angle KAP = angle BPM by subtraction property of equality

AP = BP by CPCTC

Result follows by ASA