# Prove - medial triangle, perpendicular bisector, altitude

• Dec 3rd 2009, 07:11 PM
ReneePatt
Prove - medial triangle, perpendicular bisector, altitude
The problem is to:

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

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

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

The given information is:

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

How do you write this proof out without using the diagram?
• Dec 5th 2009, 12:41 AM
red_dog
a) $C'ACB$ is a parallelogram. Then $C'A=BC$

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

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

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

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

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