# Solution to puzzle 143: Semicircle in a triangle

In ABC, side AB = 20, AC = 11, and BC = 13. Find the diameter of the semicircle inscribed in ABC, whose diameter lies on AB, and that is tangent to AC and BC.

A number of approaches to this problem are possible.

Firstly, we draw a line from C to the center of the circle O.

Then draw radii OD and OE to the points of contact of tangents CB and CA, respectively. The radii meet the tangents at right angles.

## Using Angle Bisector Theorem

By symmetry, CD = CE.

Clearly OD = OE = radius of semicircle.

Hence ODC is similar to OEC, and so OCE = OCD.

Applying the Angle Bisector Theorem to ABC, we have AC/BC = AO/BO.

Hence 11/13 = AO/(20 − AO), and so 13AO = 11(20 − AO) = 220 − 11AO.

Therefore AO = 55/6.

Applying the law of cosines (also known as the cosine rule) to ABC, we have

13^{2} = 11^{2} + 20^{2} − 2·11·20·cos A.

Hence cos A = 4/5, and so, by Pythagoras' Theorem, sin A = 3/5.

In EAO, sin A = OE/AO.

Hence OE = (3/5) · (55/6) = 11/2.

Therefore, the diameter of the semicircle is 11 units.

## Using Area of Triangles

Applying Heron's Formula to ABC,

area of ABC = [22 · (22 − 11) · (22 − 13) · (22 − 20)] = 66.

Let the radius of the semicircle be r, so that OD = OE = r.

Then area of AOC = ½ · AC · r = 11r/2.

Similarly, area of BOC = ½ · BC · r = 13r/2.

Since area of ABC = area of AOC + area of BOC, we have 66 = 12r, and hence r = 11/2.

Therefore, the diameter of the semicircle is 11 units.

Source: Original

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