In triangle ABC, D and E are mid-points of the sides BC and AC respectively. Find the length of DE. Prove that DE = 1/2AB.?

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It can be proved by converse of mid point theorem.... I cant send the image of sol... as my limit over

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Introduction:
In this problem, we are given a triangle ABC and we need to find the length of DE where D and E are midpoints of the sides BC and AC respectively. We also need to prove that DE=1/2AB.

Solution:
To find the length of DE, we can use the midpoint formula which states that the midpoint of a line segment is the average of its endpoints.

Step 1: Calculate the coordinates of D and E using the midpoint formula.

- Let the coordinates of A be (x1, y1)
- Let the coordinates of B be (x2, y2)
- Let the coordinates of C be (x3, y3)

Using the midpoint formula, we can calculate the coordinates of D and E as follows:

- Coordinates of D = ((x2 + x3)/2, (y2 + y3)/2)
- Coordinates of E = ((x1 + x3)/2, (y1 + y3)/2)

Step 2: Calculate the length of DE using the distance formula.

- Let the coordinates of D be (x4, y4)
- Let the coordinates of E be (x5, y5)

Using the distance formula, we can calculate the length of DE as follows:

- DE = sqrt((x5 - x4)^2 + (y5 - y4)^2)

Step 3: Prove that DE=1/2AB.

To prove that DE=1/2AB, we need to use the fact that D and E are midpoints of BC and AC respectively.

- Since D is the midpoint of BC, we know that BD=CD.
- Similarly, since E is the midpoint of AC, we know that AE=CE.
- Using the triangle inequality, we can say that AB+BC>AC and AC+BC>AB.
- From these inequalities, we can subtract BC from both sides to get AB>AC/2 and AB>BC/2.
- Combining these two inequalities, we get AB>(AC+BC)/2.
- But AC+BC=AB, so we can simplify the inequality to AB>AB/2.
- This simplifies to AB/2
- Similarly, we can prove that AB/2 is less than AC.
- Using the midpoint formula, we can say that DE=(AB+AC)/2.
- Substituting AB/2 for AB and AC/2 for AC, we get DE=(AB/2)+(AC/2).
- Simplifying this expression, we get DE=1/2AB.

Conclusion:
Therefore, we have calculated the length of DE and proved that DE=1/2AB using the fact that D and E are midpoints of BC and AC respectively.

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