Answer:
Segment AD is 3, and segment AE is 2.
Step-by-step explanation:
In a triangle, the line joining the mid points of two sides is parallel and half of the third sides of the triangle.
Here, ABC is a triangle,
In which,
AB = 6,
AC = 4,
D∈ AB and E∈AC
Let DE ║BC,
And, [tex]DE=\frac{1}{2}BC[/tex]
In triangles ADE and ABC,
[tex]\angle ADE\cong \angle ABC[/tex] ( Alternative interior angle theorem )
[tex]\angle AED\cong \angle ACB[/tex]
By AA similarity postulate,
[tex]\triangle ADE\sim \triangle ABC[/tex]
∵ Corresponding sides of similar triangle are in same proportion,
[tex]\implies \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}[/tex]
[tex]\frac{AD}{AB}=\frac{AE}{AC}=\frac{BC}{2BC}[/tex]
[tex]\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{2}[/tex]
[tex]\implies AD = \frac{AB}{2}\text{ and }AE =\frac{AC}{2}[/tex]
[tex]\implies AD = \frac{6}{2}=3\text{ and }AE =\frac{4}{2}=2[/tex]
Hence, the correct option would be,
Segment AD is 3, and segment AE is 2.
