Given: ABCD is an isosceles trapezoid in plane t, line BC is parallel to line AD, PF is perpendicular to t,?
(continued)PF is perpendicular to t, line PF bisects line AD, Prove triangle PAB is congruent to triangle PDC
- Ya-HooLv 48 years agoFavorite Answer
Sorry, I can't include an image because of the limitations of yahoo answers.
Since line PF bisects AD then line AP is equal to line PD. It's given that line AB and line CD are congruent as well. There are three ways to prove that triangles are congruent: Side Angle Side, Side Side Side, and Angle Side Angle. Angle APB and DPC are congruent because alternate interior angles of parallel lines are congruent. Therefore, we can conclude that triangle PAB and PDC are congruent because of Side Angle Side.
- 6 years ago
Since PF is perpendicular to t, angle PFA and PFD are right angles and are congruent. Since PF bisects AD, AF is congruent to DF. PF is congruent to itself by reflexive property. so the triangle PFA and PFD are congruent by side angle side. PA is congruent to PD by CPCTC. Since ABCD is a isosceles trapezoid, AB is congruent CD and angle BAF is congruent to angle CDF. Draw in lines BF and CF. Triangle BAF and CDF are congruent by SAS. BF is congruent to CF by CPCTC. Angle PFB and PFC are right angles and are congruent. PF is congruent itself by reflexive property. Triangle PFB is congruent to triangle PFC by SAS. PB is congruent to PC by CPCTC. Triangle PAB is congruent to triangle PDC by SSS
- 6 years ago
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