Question 156480: 1. Given: In Triangle ABC: BD is perpendicular to AC
Draw 1 conclusion from the given
2. Given triangle ABC; BD bisects Angle ABC
and BD in perpendicular to AC
Prove: AD is congruent to CD
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! not too clear on what you can conclude from number 1 since any triangle ABC can draw BD perpendicular to AC. you are forming 2 right triangles? if triangle ABC is acute, the perpendicular falls within the boundaries of triangle ABC. if triangle ABC is obtuse, the perpendicular falls outside the boundaries of triangle ABC. if triangle ABC is already a right triangle, with angle B being the right angle, then the perpendicular is already one of the sides of the triangle? i don't think the dropping of the perpendicular tells anything but would have to defer to some other expert on this one.
for number 2, however, if BD bisects angle ABC of triangle ABC, and BD is perpendicular to AC, then triangles ABD and CBD are congruant by ASA with angles ABD and CBD being one of the equal angles by bisection, and angles ADB and CDB being the other equal angles by virtue of the fact that BD is perpendicular to AC so both angles = 90 degrees. the common side DB forms the S part of the requirement because it's the same side and it's between the two equal angles so the triangles ABD and CDB are congruent and so AD is equal to DC because they are corresponding sides of congruent triangles.
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