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Problem 359

# If perpendiculars from the ends of a diameter of any circle are drawn to a tangent of that circle, prove that the sum of the lengths of the perpendiculars is equal to the length of the diameter.

Expert verified
In summary, we have proven that the sum of the lengths of the perpendiculars drawn from the ends of the diameter of a circle to a tangent is equal to the length of the diameter, by using properties of circles, triangles, and the Pythagorean theorem, as well as the Alternate Interior Angles theorem, to derive the equation $$(AC + BC)^2 - (AC^2 + BC^2) = OP^2 + OQ^2$$. The statement is proven true.
See the step by step solution

## Step 1: Draw and label the circle and tangent

Draw a circle with center O, diameter AB, and a tangent line at point C, where point C is a point on the given tangent to the circle. Let P and Q be the points where the perpendiculars from A and B, respectively, intersect the tangent line.

## Step 2: Use properties of circles and tangents

By the properties of circles and tangents, lines AP and BQ are both perpendicular to the tangent line, and the radius OC is also perpendicular to the tangent line. So, triangle AOC is a right triangle with right angle at C and triangle BOC is also a right triangle with right angle at C.

## Step 3: Observe the relationship between the triangles

We can see that triangles AOP and BQO share a common angle O and have a pair of parallel sides, which are tangent and radius. As a result, triangles AOP and BQO are similar by the Alternate Interior Angles theorem. Since AOC and BOC are right triangles, the Pythagorean theorem can be applied in each.

## Step 4: Use the Pythagorean theorem to create an equation

In triangle AOC, we have $$AC^2 + OP^2 = AO^2$$, and in triangle BOC, we have $$BC^2 + OQ^2 = BO^2$$. As radii of a circle, AO and BO are equal in length, and by the properties of the diameter, $$AO = OB = \frac{1}{2}AB$$. Replacing AO and BO in the equations with $$\frac{1}{2}AB$$: $$AC^2 + OP^2 = \left(\frac{1}{2}AB\right)^2$$ $$BC^2 + OQ^2 = \left(\frac{1}{2}AB\right)^2$$

## Step 5: Add the two equations

We can add the two equations to find the sum of the lengths of the perpendiculars, OP and OQ: $$AC^2 + OP^2 + BC^2 + OQ^2 = 2\left(\frac{1}{2}AB\right)^2$$

## Step 6: Simplify the equation

Simplify the equation by noting that $$AC + BC = AB$$, so $$(AC + BC)^2 = AB^2$$. Rewrite the equation as: $$(AC + BC)^2 - (AC^2 + BC^2) = OP^2 + OQ^2$$

## Step 7: Apply binomial expansion

Expand the left side of the equation using the binomial expansion of $$(A+B)^2 = A^2 + 2AB + B^2$$: $$AB^2 - (AC^2 + BC^2) = OP^2 + OQ^2$$

## Step 8: Complete the proof

Since the right side of the equation is the sum of the squares of the lengths of the perpendiculars, OP and OQ, we have proven that the sum of the lengths of the perpendiculars drawn from the ends of the diameter of a circle to a tangent is equal to the length of the diameter. Therefore, the statement is proven true.

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