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Q.

Prove the Converse of the mid-point Theorem following the guidelines given below:
Consider a triangle ABC with D as the midpoint of AB. Draw DE || BC to intersect AC in E. Let E1 be The midpoint of AC. we mid-point theorem to get DE1 || BC and_____.
Conclude E = E1 and hence E is the midpoint of AC.

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a

b

DE=BC

c

AE=EC

d

AD=BD

answer is A.

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Detailed Solution

Question Image

It is given that:
DE || BC
D is the midpoint of side AB and E1 is the midpoint of the side AC.
We know that E and E1 will coincide with each other. In Δ ABC and Δ ADE, we have.
∠ ABC = ∠ ADE
∠ ACB = ∠ AED
∠ BAC = ∠ DAE
∴ All the angles of Δ ABC is equal to Corresponding angles of Δ ADE.
So, by AAA rule of similarity, both the triangles are similar. Δ ABC ∼ Δ ADE
Therefore, we Can write.

…(1)

…(2)
∴ It is given that D is the midpoint of AB.
∴ AB = 2AD …(3)
Putting this value in equation 1, we get.

→ AC = 2AE
∴ E is the midpoint of AC
But it is given that E1 is the midpoint of AC Therefore, we can say that E = E1 Now using equation 2 and 3 , we get:

→ BC = 2DE
→ DE =

\frac{1}{2}

BC
Therefore, it is proved that : DE =

\frac{1}{2}

BC

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