# Concurrency Proofs

#### 12. Prove that the lines of the three altitudes of a triangle are concurrent.

To begin this proof, I will construct lines through each vertex that is perpendicular to the opposite side. Then, construct their points of intersection.

First we must prove that the altitudes of the small triangle ABC are the perpendicular bisectors of the larger triangle GHI. By construction, we know that segment BA is parallel to HI and by construction we also know that GI is parallel to BC. Since the definition of a parallelogram is that it has two sets of parallel sides, we know that BCAI must be a parallelogram. One of the properties of a parallelogram is that opposite sides are congruent, so BC=IA. By construction, we also know that AC is parallel to GB, so by difintion of parallelogram, GABC must be a parallelogram. So BC=GA by the properties of a paralleogram. Since both IA and GA are equal to BC, IA=GA.
Similarly, HC=CI and HB=BG.
So we have proven that the altitudes bisect the sides of the larger triangle, now we must prove that the altitudes are perpendicular to the sides of the larger triangle. Since GI is parallel to BC, and AE is the altitude of BC, which means it is perpendicular to BC, therefore AE must also be perpendicular to GI. Sinilarly, CD is perpendicular to HI, and BF is perpendicular to GH.
Now we have proven that the altitudes of the smaller triangle ABC are the perpendicular bisectors of the larger triangle GIH. We know that the perpendicular bisectors of a triangle are concurrent (for proof, click here), so the altitudes must also be concurrent.

#### 14. Prove that the three medians of a triangle are concurrent, and that the point of concurrence, the centroid, is two-thirds the distance from each vertex to the opposite side.

To do this proof, I will construct two triangles, Triangle1 and Triangle2.

Triangle 1:

Triangles BEF and ABC are similar by Side Angle Side Congruence theorem because they share the same angle B and their sides have a ratio of 1:2 because of the existance of midpoints. So we know that the ratio of FE to CA is 1:2. Now we can prove triangle FEG is similar to GCA by Angle Angle Angle because angle FGE=angle CGA since they are vertical angles and vertical angles of intersecting lines are congruent, and angle GFE=angle ACG and angle CAG=angle GEF because alternate interior angles of parallel lines are equal. We know that FE is parallel to CA by the converse to Euclide's theorem book 6 prop.2. Since FGE is similar to AGC, we know that all sides must have the ratio of 1:2, therefore FG:GC is 1:2.

Triangle2:

I kept the line FC the same, but I added line BD. We know all of the points F,D,E are the same because they are the midpoints, but we aren't sure that G is the same point so we rename it G'. If we can prove that G and G' are the same point, we will prove that all three lines are concurrent. We prove that triangle FDG' and triangle BCG' are similar by the same process above (Angle Angle Side similarity theorem). So we know all the sides have the ratio 1:2. Therefore FG':G'C is 1:2, so that means G'=G. Therefore, all three lines are concurrent.