Thread: Triangles

Hello, I'm having some problems doing some proofs with deductive geometry. Hope someone can help me out.
1. First one is to prove that the angle bisectors of the vertices of a triangle will all intersect at the same point.

I think I have to draw in the first 2 angle bisectors (AF and EB), then for the third one, I will have to draw a line(DC) through the point where the first 2 bisectors intersect. Then I have to prove that angle ACD = DCB to prove that it intersects at the middle.

2. The second one seems similar to the first one. Instead of angle bisectors, this time its the altitudes. I have to prove that the 3 altitudes of a triangle will intersect at the same point.

Similar to the first one, I think I have to put in the first 2 altitudes and then for the third one, I have to prove that it is 90 degrees from the base to confirm that it intersects at the same point as the other two.

Originally Posted by ilovemaths321

Hello, I'm having some problems doing some proofs with deductive geometry. Hope someone can help me out.
1. First one is to prove that the angle bisectors of the vertices of a triangle will all intersect at the same point.

I think I have to draw in the first 2 angle bisectors (AF and EB), then for the third one, I will have to draw a line(DC) through the point where the first 2 bisectors intersect. Then I have to prove that angle ACD = DCB to prove that it intersects at the middle.

2. The second one seems similar to the first one. Instead of angle bisectors, this time its the altitudes. I have to prove that the 3 altitudes of a triangle will intersect at the same point.

Similar to the first one, I think I have to put in the first 2 altitudes and then for the third one, I have to prove that it is 90 degrees from the base to confirm that it intersects at the same point as the other two.

For 1, read here:

Triangle Proofs

i found this at another forum:PROVE: The altitude of a triangle are concurrent.

Originally Posted by anonimnystefy

i found this at another forum:PROVE: The altitude of a triangle are concurrent.

Hmmm, but I wasn't given a.BC=0 and such, like in that post.

it is always the case when two vectors are perpendicular.it is because then the angle beteen them is 90 degrees or pi/2 radians,so the dot product becomes:
a.BC=|a|.|BC|cos pi/2=|a|.|BC|.0=0
that was just a hint to help you do it by vectors.

Originally Posted by anonimnystefy

it is always the case when two vectors are perpendicular.it is because then the angle beteen them is 90 degrees or pi/2 radians,so the dot product becomes:
a.BC=|a|.|BC|cos pi/2=|a|.|BC|.0=0
that was just a hint to help you do it by vectors.

I think I have to prove it using deductive geometry only, and I haven't learned the vectors/dot products yet.

oh ok

than i'm not sure i can help you.