In a ABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.

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Solution

is given with E and F as the mid points of sides AB and AC.

Also, intersecting EF at Q.

We need to prove that

In , E and F are the mid-points of AB and AC respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, we get:

Since, Q lies on EF.

Therefore,

This means,

Q is the mid-point of AP.

Thus, (Because, F is the mid point of AC and)

Hence proved.

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Similar questions

In a $Δ A B C,$ E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.

In triangle ABC, E and F are the mid points of the sides AC and AB respectively. The altitude AP to BC intersects EF at Q. Then

Assertion: In $Δ A B C$ , E and F are the mid-points of AC and AB respectively. The altitude AP at BC intersect and FE at Q. Then AQ = QP.
Reason: Q is the mid - point of AP.
Which of the following is correct?

Q. In

Δ A B C

, M and N are the mid-points of AB and AC respectively. The altitude AP to BC interest MN at O.

Prove that

A O = O P

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