# Prove - medial triangle, perpendicular bisector, altitude

• Dec 3rd 2009, 06:11 PM
ReneePatt
Prove - medial triangle, perpendicular bisector, altitude
The problem is to:

(1) Prove that \$\displaystyle \Delta\$ABC is the medial triangle for \$\displaystyle \Delta\$A'B'C'.

(2) Prove that the perpendicular bisector of a side of \$\displaystyle \Delta\$A'B'C' is an altitude for \$\displaystyle \Delta\$ABC.

(3) Prove that the three altitudes of any triangle are concurrent.

The given information is:

Given a triangle \$\displaystyle \Delta\$ABC, draw a line through each vertex that is parallel to the opposite side. The three lines intersect to form a new triangle \$\displaystyle \Delta\$A'B'C'. The diagram can not be used as the proof.

How do you write this proof out without using the diagram?
• Dec 4th 2009, 11:41 PM
red_dog
a) \$\displaystyle C'ACB\$ is a parallelogram. Then \$\displaystyle C'A=BC\$

\$\displaystyle AB'CB\$ is also a parallelogram, then \$\displaystyle AB'=BC\$.

Then \$\displaystyle C'A=AB'\$ and \$\displaystyle A\$ is the midpoint of \$\displaystyle B'C'\$.

Do the same for the others sides of the triangle \$\displaystyle A'B'C'\$

b) Let \$\displaystyle AM\perp BC\$. But \$\displaystyle BC\parallel B'C'\Rightarrow AM\perp B'C'\$. Then \$\displaystyle AM\$ is the perpendicular bisector of \$\displaystyle B'C'\$.

c) The perpendicular bisectors of the sides of the triangle \$\displaystyle A'B'C'\$ are concurrent, so the altitudes of the triangle \$\displaystyle ABC\$ are concurrent.