Math 8 Chapter 1 Lesson 4: Midsegment of a triangle, of a trapezoid
1. Summary of theory
1.1. Midsegment of the triangle:
– Define: The midsegment of a triangle is the line segment joining the midpoints of the two sides of the triangle.
Eg: In the figure above, we call DE the median of triangle ABC.
– The definitions:

The line passing through the midpoint of one side of the triangle and parallel to the second side passes through the midpoint of the third side.

The midsegment of a triangle is parallel to the third side and is half of that side.
1.2. The median of the trapezoid:
– Define: The midsegment of a trapezoid is the line segment joining the midpoints of the two sides of the trapezoid.
Eg: In the figure above, we call EF the median of trapezoid ABCD.
– The definitions:

The line passing through the midpoint of one side of the trapezoid and a river with two bases passes through the midpoint of the second side.

The median of a trapezoid is parallel to the bases and is half the length of the sum of the two bases.
2. Illustrated exercise
Question 1: Let ABC be a triangle with D, E, F being the midpoints of the names AB, AC, BC, respectively. Let G be the midpoint of AF. Prove that D,G,E are collinear and G is the midpoint of DE.
Solution guide
 Prove that D, G, E are collinear
Consider triangle ABF with:
D is mid point AB
G is the midpoint AF
⇒DG is the median of triangle ABF
\( \Rightarrow DG\parallel BF\) and \(DG = \frac{1}{2}BF\)
Consider triangle AFC with:
G is the midpoint AF
E is mid point AC
⇒GE is the median of triangle AFC
\( \Rightarrow GE\parallel FC\) and \(GE = \frac{1}{2}FC\)
We have: \( DG\parallel BF\) and \( GE\parallel FC\) ⇒D, G, E collinear (Euclidean axiom)
 Prove that G is the midpoint of DE .
We have: \(DG = \frac{1}{2}BF\) and \(GE = \frac{1}{2}FC\)
Where BF=CF (F is the midpoint of BC)
⇒DG=GE
Where D, E, G are collinear
⇒G is mid point of DE
Verse 2: Let ABC be a triangle with BD and CE being the medians intersecting at G, let I, K be the midpoints of GB and GC, respectively. Prove that \(DE\parallel IK\) and DE=IK
Solution guide
Consider triangle ABC with:
E is mid point AB
D is mid point AC
⇒DE is the median of triangle ABC.
\( \Rightarrow DE\parallel BC\) and \(DE = \frac{1}{2}BC\)
Similar to triangle GBC we also have:
I is the midpoint GB
K is the midpoint of GC
⇒IK is the median of triangle GBC.
\( \Rightarrow IK\parallel BC\) and \(IK = \frac{1}{2}BC\)
\( \Rightarrow IK\parallel DE\) (same parallel to BC ) and IK=DE (both equal to half BC)
Question 3: Let ABCD be a trapezoid with AB, CD being the two bases and AB < CD. Let E and F be the midpoints of BD and AC respectively. Prove that: \(EF = \frac{{CD – AB}}{2}\)
Solution guide
Let G and H be the midpoints of AD and BC respectively.
Consider triangle ADC with:
G is mid point AD
F is mid point AC
⇒GF is the median of triangle ADC
\( \Rightarrow GF\parallel DC\) and \(GF = \frac{1}{2}CD\)
Similar proof with triangle BCD we also have: EH is the median of triangle BCD
\( \Rightarrow EH\parallel CD\) and \(EH = \frac{1}{2}CD\)
We have \(GF\parallel DC\) and \(EH\parallel CD\) ⇒E,F,G,H are collinear.
Considering triangle ABD, it is easy to see that GE is the median of triangle ABD, so \(GE = \frac{1}{2}AB\)
Similar to triangle ABC, we can also prove \(FH = \frac{1}{2}AB\)
On the other hand, we have that GH is the median of trapezoid ABCD, so \(GH = \frac{{AB + CD}}{2}\)
We have
\(\begin{array}{l} GH = GE + EF + FH\\ \,\,\,\,\,\,\,\, = \frac{1}{2}AB + EF + \frac {1}{2}AB\\ \,\,\,\,\,\,\,\,\,= AB + EF = \frac{{AB + CD}}{2} \end{array}\ )
\( \Rightarrow EF = \frac{{CD – AB}}{2}\) (something to be proved)
3.1. Essay exercises
Question 1: Given triangle ABC, point D belongs to side AC such that AD = 1/2 DC, Let M be mid point of BC, I be intersection of BD and AM. Prove: AI = IM.
Verse 2: Trapezoid ABCD has bases AB and CD. Let E, F, I be the midpoints of AD, BC, AC respectively. Prove that the three points E, F, I win the row.
Question 3: Given quadrilateral ABCD. Let E, F, I be the middle of AD, BC, AC, respectively. Prove that: EI//CD, IF//AB
Question 4: Given trapezoid ABCD (AB // CD), M is mid point of AD, N is mid point of BC. Let I, K, respectively, be the intersection of MN with BD, AC. Let AB = 6cm, CD = l4cm. Calculate the lengths MI, IK, and KN.
3.2. Multiple choice exercises
Question 1: Given a triangle ABC, D is the midpoint of AB and E is the midpoint of AC, which statement is false?
A. DE is the median of triangle ABC
B. DE is parallel to BC
C. DECB is an isosceles trapezoid
D. DE is half the length of BC
Verse 2: Let ABC be a triangle with D, E being the midpoints of AB, AC and DE = 4 cm, respectively. Know the height AH = 6 cm . What is the area of triangle ABC?
A. \(S = 24c{m^2}\,\)
B. \(\,\,S = 12c{m^2}\,\,\)
C. \(S = 48c{m^2}\)
D. \(S = 32c{m^2}\)
Question 3: Choose the correct statement
A. The median of a trapezoid is the line segment connecting the midpoints of the two sides of the trapezoid
B. The midsegment of a trapezoid is the line segment joining the midpoints of the two opposite sides of the trapezoid
C. The median of a trapezoid is parallel to the two bases and is equal to the sum of the two bases
D. A trapezoid can have one or more moving averages
Question 4: With a, b, h being the large base, small base and height of a trapezoid, respectively, the formula for calculating the area of a trapezoid is?
A. \(S = \left( {a + b} \right).h\)
B. \(S = \frac{1}{2}\left( {a + b} \right).h\)
C. \(S = \frac{1}{3}\left( {a + b} \right).h\)
D. \(S = \frac{1}{4}\left( {a + b} \right).h\)
4. Conclusion
Through this lesson, you should know the following:
 Determine the midsegment of the triangle of a trapezoid.
 Remember the properties of the midsegment of a trapezoid.
 Apply knowledge to solve related problems.
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