A segment is bounded by two endpoints.
The two segments can have up to two common points
Assume the line segments are AB and CD where the length of AB is greater than the length of CD.
<u>The possibilities are:</u>
- <em>A point of segment CD lies on segment AB</em>
- <em>Both points of segment CD lie on segment AB.</em>
<em />
See attachment for both possibilities.
Hence, it is possible for the two segments to have two common points.
Read more about line segments at:
brainly.com/question/18983323
Answer:
rhombus
Step-by-step explanation:
A quadrilateral with equal-length sides is a <em>rhombus</em>.
_____
A square is a special case of a rhombus.
Answer:
6.17
Step-by-step explanation:
Answer:
Step-by-step explanation:
5+1+3+8+2=19
3/19 = 16%
We have that AB || DC.
By a similar argument used to prove that AEB ≅ CED,we can show that (AED) ≅ CEB by (SAS) . So, ∠CAD ≅ ∠ (ACB) by CPCTC. Therefore, AD || BC by the converse of the (
ALTERNATE INTERIOR ANGLES) theorem. Since both pair of opposite sides are parallel, quadrilateral ABCD is a parallelogram
1. AED
2. SAS
3. ACB
4. ALTERNATE INTERIOR ANGLES
