1. Parallel Line Segments and the Midpoint Theorem
Slideshow 35, Mathematics
Mr. Richard Sasaki

2. Objectives
Understand the influence of parallel line segments on triangle structures
Be able to calculate missing line segments for structures with parallel line segments
Prove triangles are similar with parallel line segments
Use the midpoint theorem

3. Parallel Line Segments
Have a look at the diagram below. 𝐴
What can we say about 𝐵𝐶 and 𝐷𝐸 ?
They’re parallel.
( 𝐵𝐶 ∥ 𝐷𝐸 )
80 𝑜
As they are parallel, the triangles must be 𝐶 𝐵
similar
(∆𝐴𝐵𝐶~∆𝐴𝐷𝐸)
80 𝑜 𝐷 𝐸
Why would a pair of parallel line segments imply similarity?
𝑥 𝑜
𝑦 𝑜
We’d know the angles either side are equal, so the triangles are similar by test AA.
𝑥 𝑜
𝑦 𝑜

4. Parallel Line Segments
If a pair of parallel line segments exist and by test AA, similarity exists, we can assume all edges are in the same proportions (SSS). 𝐴
Let’s make an equation with ratios for the given line segments.
12 𝑐𝑚
𝑥 𝑐𝑚
= = ①
𝐴𝐵 : 𝐴𝐷
𝐴𝐶 : 𝐴𝐸
𝐵𝐶 : 𝐷𝐸 𝐶 𝐵 ②
Also…
𝐴𝐵 : 𝐵𝐷 =
𝐴𝐶 : 𝐶𝐸
6 𝑐𝑚
8 𝑐𝑚
The opposite also applies too. 𝐷 𝐸
You may use ① or ② for this but if 𝐵𝐶 or 𝐷𝐸 are included, we can only use ①.
12 6 = 𝑥 8 ⇒
Let’s use ②.
12:6=𝑥:8⇒
𝑥=16

5. Answers – Easy (Q1 – 2) 𝑥= 15 2 𝑥= 45 7
𝐴𝐵 : 𝐴𝐷 =4:10⇒ Value of Ratio = 2 5
𝐴𝐵 : 𝐵𝐷 =4:6⇒ Value of Ratio = 2 3
No, they aren’t the same.

6. Answers – Easy (Question 3)
Consider two triangles ∆𝐴𝐵𝐶 and ∆𝐷𝐸𝐶 where D and E are points on 𝐴𝐶 and 𝐵𝐶 respectively, forming 𝐷𝐸 .
∠𝐴𝐶𝐵=∠𝐷𝐶𝐸 as they are common in both triangles. 𝐴𝑙𝑠𝑜, 𝐴𝐷 : 𝐴𝐶 = 𝐵𝐸 : 𝐵𝐶 as stated.
∴, by 𝑆𝐴𝑆, ∆𝐴𝐵𝐶~∆𝐷𝐸𝐶.
∆𝐴𝐵𝐶~∆𝐷𝐸𝐶 where 𝐷 and 𝐸 are points on 𝐴𝐶 and 𝐵𝐶 respectively and ∠𝐴𝐶𝐵 (∠𝐷𝐶𝐸) is common in both. As 𝐴𝐵 and 𝐷𝐸 are corresponding edges, corresponding angles either side of these edges are equal. Hence, ∠𝐷𝐸𝐶=∠𝐴𝐵𝐶 (and similarly, ∠𝐸𝐷𝐶=∠𝐵𝐴𝐶). As both edges are corresponding and corresponding pairs of angles are equal, 𝐴𝐵 ∥ 𝐷𝐸 .

7. 𝑥=7, 𝑦=
∆𝐴𝐵𝐶≅∆𝐴𝐷𝐸
No. If they are not parallel, the angles would be different, therefore by AA, the triangles can’t be similar unless ∠𝐴𝐹𝐺=∠𝐴𝐸𝐷.
𝑎=6, 𝑏= 16 5
𝑥= 21 5 , 𝑦= 50 3
𝑐= , 𝑑= 30 7

8. The Midpoint Theorem
What can we say about a line segment built from two midpoints of line segments on a triangle? 𝐴
Consider some ∆𝐴𝐵𝐶.
Mid-segment
Label the midpoints on 𝐴𝐵 and 𝐴𝐶 , 𝐷 and 𝐸 respectively.
𝑎 𝑜 𝐷 𝐸
Construct 𝐷𝐸 .
As 𝐷 and 𝐸 are midpoints, 𝐴𝐷 = 𝐷𝐵 and 𝐴𝐸 = 𝐸𝐶 . 𝐵 𝐶
As 𝐴𝐵 = 𝐴𝐷 + 𝐷𝐵 , we can say 𝐴𝐵 =2 𝐴𝐷 .
We can also say 𝐴𝐶 =2 𝐴𝐸 .
Consider the angle at Vertex A, ∠𝑎.
As ∠𝑎 is common in both ∆𝐴𝐷𝐸 and ∆𝐴𝐵𝐶, and 𝐴𝐵 : 𝐴𝐷 = 𝐴𝐶 : 𝐴𝐸 , by test SAS, ∆𝐴𝐷𝐸~∆𝐴𝐵𝐶.

9. The Midpoint Theorem 𝐴
∆𝐴𝐷𝐸~∆𝐴𝐵𝐶 where 𝐴𝐵 =2 𝐴𝐷 and 𝐴𝐶 =2 𝐴𝐸 are true. Therefore 𝐵𝐶 =2 𝐷𝐸 (corresponding line segments).
𝑎 𝑜 𝐷 𝐸
As ∆𝐴𝐷𝐸~∆𝐴𝐵𝐶, ∠𝐴𝐷𝐸=∠𝐴𝐵𝐶 due to being corresponding angles. ∴ 𝐷𝐸 ∥ 𝐵𝐶 . 𝐵 𝐶
A segment that connects the midpoints of any two edges is parallel to and half the length of the third edge.
Also, we can use the midpoint theorem with trapeziums and parallelograms by splitting them into triangles.

10. 𝑥=4, 𝑦=6,𝑧=5,
𝑎=56
The area of the triangle is a third of the area of the trapezium.
𝐸𝐹 =3.5 𝑚𝑚 𝐹
𝐹𝐺 =9 𝑚𝑚
𝑥=25, 𝑦=21, 𝑧=15

11. Answers – Part 2, Question 1𝐴
Consider some ∆𝐴𝐵𝐶 with mid-segment 𝐷𝐸 . As 𝐷 and 𝐸 are midpoints on 𝐴𝐵 and 𝐴𝐶 respectively, 𝐴𝐵 = 𝐴𝐷 + 𝐷𝐵 where 𝐴𝐷 = 𝐷𝐵 .
𝑎 𝑜 𝐷 𝐸
∴ 𝐴𝐵 =2 𝐴𝐷 and similarly, 𝐴𝐶 =2 𝐴𝐸 . ∴ 𝐴𝐵 : 𝐴𝐷 = 𝐴𝐶 : 𝐴𝐸 . Also, consider ∠𝑎 as shown. ∠𝑎 is common in both. ∴ by test SAS, ∆𝐴𝐷𝐸~∆𝐴𝐵𝐶. 𝐵 𝐶
∆𝐴𝐷𝐸~∆𝐴𝐵𝐶 where 𝐴𝐵 =2 𝐴𝐷 and 𝐴𝐶 =2 𝐴𝐸 are true. Therefore 𝐵𝐶 =2 𝐷𝐸 (corresponding line segments). As ∆𝐴𝐷𝐸~∆𝐴𝐵𝐶, ∠𝐴𝐷𝐸=∠𝐴𝐵𝐶 as they are corresponding angles. As they share ∠𝑎, 𝐷𝐸 ∥ 𝐵𝐶 .

12. Consider some 𝐵𝐷 . For ∆𝐴𝐵𝐷, E and H are midpoints of 𝐵𝐴 and 𝐴𝐷 respectively.
∴ 𝐸𝐻 ∥ 𝐵𝐷 and 𝐸𝐻 = 𝐵𝐷 2 . Also similarly for ∆𝐶𝐷𝐵, 𝐹𝐺 ∥ 𝐵𝐷 and 𝐹𝐺 = 𝐵𝐷 2 .
∴ 𝐸𝐻 ∥ 𝐹𝐺 and 𝐸𝐻 = 𝐹𝐺 . As one pair of edges are congruent and parallel, quadrilateral EFGH is a parallelogram.
Two kites and two parallelograms
Both pairs of shapes are congruent
The two kites are similar to the main one.
Scale Factor: 1 2
The area is a quarter of the size of the original.