Semicircle Thales Theorem Questions with Solutions

Problem Statement: In the figure below, \(A\), \(B\), and \(D\) are points on the circle and \(AB\) passes through the center \(O\) of the circle whose radius \(r = 10\).

1. Find all angles and sides of triangle \(ADB\).
2. Find the size of angle \(\angle AOD\).

Solution:

1. According to the semicircle theorem above, \(\angle ADB = 90^\circ\) and therefore \(\angle DBA = 90^\circ - 63^\circ = 27^\circ\).
   \(AB = 2r = 20\).
   \(ADB\) is a right triangle, and therefore: \[ \cos 63^\circ = \frac{AD}{AB} = \frac{AD}{20} \quad \text{and} \quad \sin 63^\circ = \frac{DB}{AB} = \frac{DB}{20} \] This gives: \[ AD = 20 \cos 63^\circ \quad \text{and} \quad DB = 20 \sin 63^\circ \]
2. \(\angle AOD\) is a central angle intercepting the same arc as the inscribed angle \(\angle DBA\), hence: \[ \angle AOD = 2 \angle DBA = 54^\circ \]

Problem Statement: In the figure below, \(A\), \(B\), \(C\), and \(D\) are points on a circle and \(AC\) is its diameter.

1. Find \(x\) given the lengths of segments \(DA = x - 4\), \(DC = x + 2\), \(BA = x - 2\), and \(BC = x + 1\).
2. Find the radius of the circle.
3. Find the area of the quadrilateral \(ABCD\).

Solution:

1. According to the semicircle theorem, \(CAB\) and \(CAD\) are right triangles with a common hypotenuse \(AC\). Use the Pythagorean theorem to write: \[ (x - 4)^2 + (x + 2)^2 = AC^2 \quad \text{(I)} \] and \[ (x - 2)^2 + (x + 1)^2 = AC^2 \quad \text{(II)} \] Combining (I) and (II), we obtain the equation: \[ (x - 4)^2 + (x + 2)^2 = (x - 2)^2 + (x + 1)^2 \] Expanding: \[ 2x^2 - 4x + 20 = 2x^2 - 2x + 5 \] Simplifying: \[ -2x + 15 = 0 \implies x = \frac{15}{2} \]
2. The radius \(r\) of the circle is equal to half the diameter \(AC\). Substitute \(x\) into (I): \[ AC^2 = \left(\frac{15}{2} - 4\right)^2 + \left(\frac{15}{2} + 2\right)^2 = \frac{205}{2} \] \[ AC = \sqrt{\frac{205}{2}} \implies r = \frac{1}{2} \sqrt{\frac{205}{2}} \]
3. The area \(A_1\) of right triangle \(CAB\) is: \[ A_1 = \frac{1}{2} \times BA \times BC = \frac{1}{2} (x - 2)(x + 1) \] The area \(A_2\) of right triangle \(CAD\) is: \[ A_2 = \frac{1}{2} \times DA \times DC = \frac{1}{2} (x - 4)(x + 2) \] The area \(A\) of quadrilateral \(ABCD\) is the sum of \(A_1\) and \(A_2\): \[ A = A_1 + A_2 = \frac{1}{2} \left((x - 2)(x + 1) + (x - 4)(x + 2)\right) \] Substitute \(x = \frac{15}{2}\): \[ A = \frac{1}{2} \left(\left(\frac{15}{2} - 2\right)\left(\frac{15}{2} + 1\right) + \left(\frac{15}{2} - 4\right)\left(\frac{15}{2} + 2\right)\right) = 40 \, \text{square units} \]

Problem Statement: In the figure below, \(A\), \(B\), and \(C\) are points on a semicircle whose diameter is \(AB = 50\). The sides \(CA\) and \(CB\) of triangle \(ABC\) are congruent.

1. Find the lengths of \(CA\) and \(CB\).
2. Find the area of the shaded (blue) surface.

Solution:

1. According to the semicircle theorem, \(ACB\) is a right triangle. Using the Pythagorean theorem: \[ CA^2 + CB^2 = AB^2 \] Given \(AB = 50\) and \(CA = CB\): \[ 2CA^2 = 50^2 \implies CA = \frac{50}{\sqrt{2}} = CB \]
2. Let \(A_s\) be the area of the semicircle, \(A_t\) the area of triangle \(ACB\), and \(r\) the radius of the semicircle. \[ A_s = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \left(\frac{AB}{2}\right)^2 = \frac{25^2}{2} \pi \] \[ A_t = \frac{1}{2} \times CA \times CB = \frac{1}{2} \times \frac{50^2}{2} \] The shaded area \(A\) is the area of the semicircle minus the area of the triangle: \[ A = \frac{25^2}{2} \pi - \frac{1}{2} \times \frac{50^2}{2} = \frac{625\pi}{2} - 625 \approx 356.75 \, \text{square units} \]

Problem Statement: \(C\) is a point on the circle whose diameter is \(AB = 100\) and center \(O\).

1. Find angles \(\angle COB\) and \(\angle COA\).
2. Find the areas of triangles \(COA\) and \(COB\).
3. Find the lengths of \(CA\) and \(CB\).
4. Find the area of triangle \(ACB\).
5. Compare the area of triangle \(ACB\) to the sum of the areas found in part 2.

Solution:

1. Points \(A\), \(C\), and \(B\) are on the circle with center \(O\), so \(OA = OB = OC = 50\). Triangles \(AOC\) and \(COB\) are isosceles.
   Given \(\angle OCB = \angle CBO = 62^\circ\), this gives: \[ \angle COB = 180^\circ - 62^\circ - 62^\circ = 56^\circ \] Since \(\angle COB\) and \(\angle COA\) form a straight angle: \[ \angle COA = 180^\circ - \angle COB = 124^\circ \]
2. Using the sine rule formula for area: \[ A_1 = \frac{1}{2} \sin(\angle COA) \times OC \times OA = \frac{1}{2} \sin(124^\circ) \times 50 \times 50 = 1250 \sin(124^\circ) \approx 1036.30 \, \text{square units} \] Similarly, for triangle \(COB\): \[ A_2 = \frac{1}{2} \sin(\angle COB) \times OC \times OB = \frac{1}{2} \sin(56^\circ) \times 50 \times 50 = 1250 \sin(56^\circ) \approx 1036.30 \, \text{square units} \]
3. According to the semicircle theorem, \(ACB\) is a right triangle with \(AB\) as hypotenuse: \[ \sin(62^\circ) = \frac{CA}{AB} \implies CA = 100 \sin(62^\circ) \] \[ \cos(62^\circ) = \frac{CB}{AB} \implies CB = 100 \cos(62^\circ) \]
4. The area \(A\) of right triangle \(ACB\) is: \[ A = \frac{1}{2} \times CA \times CB = \frac{1}{2} (100)^2 \sin(62^\circ) \cos(62^\circ) \approx 2072.59 \, \text{square units} \]
5. The sum of the areas in part 2 is: \[ A_1 + A_2 = 1036.30 + 1036.30 \approx 2072.60 \] This is extremely close to \(A \approx 2072.59\), with the minute difference arising from rounding errors. In fact, the sum of the areas of triangles \(COA\) and \(COB\) equals the area of triangle \(ACB\).