GEOM 3| Lesson 3 | Try This! (Midsegment Theorem)

 

Try This!

The Midsegment Theorem:   Segment joining the midpoints of two sides of the triangle is parallel to the third side and is \(\frac{1}{2}\) its length.

As a consequence to this theorem, \(ΔABC ≈ΔADC\) and \(ΔABC ≈ΔFEC\).

Is \(ΔABC ≈ΔEFD\)?

What other similarities or congruences do you find in the figure?

  1. In the figure above, \(BC = 10, AB = 6, AC = 8\), find \(XY, YZ\) and \( XZ\).
  2.  \(BC = 12, YZ = 4,\) and \( AC = 10\), find \(XY, XZ \) and \(AB\).

Did you find

  1.  \(XY = 5, YZ = 3\) and \(XZ =4\)
  2. \(XY = 6, XZ = 5\) and \(AB=8\)

Go to Practice 2