# The diagram shows a first class lever. How far down will the effort side have to move to lift the load 1 m?

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Hello!

The answer is obviously **2m**, because the effort side is twice as long as the load side. But it is better to prove this.

Please look at the picture.

The load goes from the point B to point B', our hands go from A to A'.

The height to which B' is lifted is B'C, perpendicular to the ground.

The height down to which A' goes is DA', perpendicular to the line, parallel to the ground.

Now consider triangles A'DO and B'CO. They are similar because have two pairs of equal angles. Angles A'OD and B'OC are equal as vertical angles. Angles A'DO and B'CO are equal as the complementary to the alternate interior angles ODA and OCB.

So A'DO and B'CO are similar. Also A'O=4m and B'O=2m, A'O/B'O=2. Therefore

`(A'D)/(B'C) = 2,`

and because B'C is known to be 1m, A'D=**2m**, QED.