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In the trapezoid ABCD ( AB ∥ CD ) point M∈ AD , so that AM:MD=3:5. Line l ∥ AB and going trough point M intersects diagonal AC and leg BC at points P and N respectively. Find BC:BN

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In the given trapezoid ABCD, point M is located on the side AD such that the ratio of AM to MD is 3:5. Line l is parallel to side AB and passes through point M, intersecting diagonal AC at point P and leg BC at point N. We need to find the ratio of BC to BN.

Let's start by drawing the trapezoid and labeling the given information:

Step 1: Draw trapezoid ABCD, with AB parallel to CD.

Step 2: Mark point M on side AD, such that AM:MD = 3:5.

Step 3: Draw line l parallel to AB, passing through point M, and intersecting diagonal AC at point P and leg BC at point N.

Step 4: Label the length of AB as x.

Step 5: Using the properties of similar triangles, we can determine the ratio BC:BN as follows:

Step 6: Triangle AMN is similar to triangle BNC due to the corresponding angles formed by parallel lines and the transversal line l. This gives us the proportional relationship:

AM/BN = AN/BC = MN/NC

Step 7: We know that AM:MD = 3:5, so we can find the lengths of AM and MD:

AM = (3/(3+5))*AD

MD = (5/(3+5))*AD

Step 8: From the similarity of triangles AMN and BNC, we can write:

AM/BN = AN/BC

Substituting the values of AM and MD, we get:

((3/(3+5))*AD)/BN = ((5/(3+5))*AD)/BC

Simplifying this equation, we can cancel out AD:

(3/(3+5))/BN = (5/(3+5))/BC

Cross multiplying, we get:

3BC = 5BN

Dividing both sides by BN, we get:

BC/BN = 5/3

final answer is The ratio of BC to BN is 5:3.

User Alfonso Nishikawa
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