If point E is the midpoint of segment AC and point D is the midpoint of segment BC, which expression represents the value of s? triangle CAB, point E is on segment AC between points A and C and point D is on segment BC between points B and C, creating segment ED, CE equals p, EA equals r, CD equals q, DB equals t, ED equals s, and AB equals u s equals p over q s = one halfu s equals q over p s = 2u

The side ED is a midsegment of the triangle ΔABC, therefore, the length of

ED is half the length of AB.

[tex]\displaystyle The \ expression \ that \ represent \ the \ value \ of \ s \ is; \ s = \frac{1}{2} \cdot u[/tex]

Reasons:

The given parameters are;

In ΔABC, point E is the midpoint of AC

The midpoint of BC is the point D

Segment ED = s

Segment CE = p

Segment EA = r

Segment CD = q

Segment DB = t

Segment ED = s

Segment AB = u

Required;

The expression that represents the value of s

Solution:

CE = 0.5 × AC Definition of midpoint

CD = 0.5 × CB Definition of midpoint

Therefore;

Therefore, we have;

[tex]\displaystyle \frac{CE}{AC} = \mathbf{\frac{CD}{CB}} = 0.5[/tex]

Therefore, given that ∠C ≅ ∠C, by reflexive property, we have;

ΔABC is similar to ΔCDE by Side-Angle-Side similarity

Which gives;

[tex]\displaystyle \frac{CE}{AC} = \frac{CD}{CB} = \mathbf{\frac{ED}{AB} } = 0.5 = \frac{1}{2}[/tex]

ED = s and AB = u which gives;

[tex]\displaystyle \frac{ED}{AB} =\frac{s}{u} = 0.5 = \frac{1}{2}[/tex]

[tex]\displaystyle \frac{s}{u} = \frac{1}{2}[/tex]

Which gives;

[tex]\displaystyle s = \mathbf{ \frac{1}{2} \times u}[/tex]

(The above relationship is given by the midsegment theorem)

Therefore;

The expression that represents the value of s is; s = one half u

Learn more about midsegment of a triangle here:

https://brainly.com/question/11759880

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