# Semicircle Thales Theorem Questions with Solutions

The semicircle (or Thales ) theorem is presented along with questions and their detailed solutions.

## Semicircle Thales Theorem Statement

The semicircle (Thales) theorem [1] states that for a triangle inscribed in a semicircle of diameter $AB$, as shown in the figure below, angle $AOB$ has a size of $90^{\circ}$.

## Question With Solutions

Note that none of the figures below is drawn to scales.
Question 1
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 $\angle DBA = 90^{\circ} - 63^{\circ} = 27^{\circ}$
$AB = 2 r = 20$ ,
$ADB$ is a right triangle and therefore $\quad \cos 63^{\circ} = \dfrac{AD}{AB} = \dfrac{AD}{20}$ and $\quad \sin 63^{\circ} = \dfrac{DB}{AB} = \dfrac{DB}{20}$
The above gives: $\quad AD = 20 \cos 63^{\circ}$ 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 $\angle AOD = 2 \angle DBA = 54^{\circ}$

Question 2
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 $\quad DA = x-4 , DC = x+2, BA = x-2$ and $BC = x + 1$.
2) Find the radius of the circle.
2) Find the area of the quadrilateral $ABCD$.

Solution
1)
According to the semicircle theorem above, $CAB$ and $CAD$ are right triangles with a common hypotenuse. Use the Pythagorean theorem to write:
$(x-4)^2 + (x+2)^2 = AC^2$     (I)
and   $(x-2)^2 + (x+1)^2 = AC^2$     (II)
Combining (I) and (II) above, we obtain the equation: $\quad (x-4)^2 + (x+2)^2 = (x-2)^2 + (x+1)^2$
Expand: $2x^2-4x+20 = 2x^2-2x+5$
Group like terms and simplify to rewrite the above equation as: $\quad -2x+15 = 0$
Solve for $x$: $\quad x = \dfrac{15}{2}$
2)
The radius $r$ of the circle is equal to half the diameter $AC$
Substitute $x$ by its value in (I) to write: $\quad AC^2 = (\dfrac{15}{2}-4)^2 + (\dfrac{15}{2}+2)^2 = \dfrac{205}{2}$
$AC = \sqrt {\dfrac{205}{2}}$ , hence $\quad r = \dfrac{1}{2} \sqrt {\dfrac{205}{2}}$
3)
The area $A_1$ of the right triangle triangle $CAB$ is given by: $\quad A_1 = \dfrac{1}{2} \times BA \times BC = \dfrac{1}{2} (x-2)(x+1)$
The area $A_2$ of the right triangle triangle $CAD$ is given by: $\quad A_2 = \dfrac{1}{2} \times DA \times DC = \dfrac{1}{2} (x-4)(x+2)$
The area $A$ of the quadrilateral $ABCD$ is the sum of the areas $A_1$ and $A_2$ of triangles $CAB$ and $CAD$
$A = A_1 + A_2 = \dfrac{1}{2} ( (x-2)(x+1) + (x-4)(x+2) )$
Substitute $x$ by its value found above: $A = \dfrac{1}{2} ( (\dfrac{15}{2}-2)(\dfrac{15}{2}+1) + (\dfrac{15}{2}-4)(\dfrac{15}{2}+2) ) = 40$ square units.

Question 3
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 above, $ACB$ is a right triangle. Use the Pythagorean theorem to write: $\quad CA^2 + CB^2 = AB^2$
Given that $AB = 50$ and $CA$ and $CB$ are congruent, the above gives the equation: $\quad 2 CA^2 = 50^2$
Solve for $CA$ to obtain: $\quad CA = \dfrac{50}{\sqrt 2} = CB$
2)
Let $A_s$ be the area of the semicircle, $A_t$ the area of triangle $ACB$ and $r$ the radius of the semicircle.
$A_s = \dfrac{1}{2} \pi r^2 = \dfrac{1}{2} \pi \left(\dfrac{AB}{2}\right)^2 = \dfrac{25^2}{2} \pi$
$A_t = \dfrac{1}{2} \times CA \times CB = \dfrac{1}{2} \times \dfrac{50^2}{ 2 }$
The shaded area $A$ is the area of the triangle $A_t$ subtracted from the area of the semicircle $A_s$.
$A = \dfrac{25^2}{2} \pi - \dfrac{1}{2} \times \dfrac{50^2}{ 2 } = \dfrac{625\pi }{2}-625 \approx 356.75$ square units.

Question 4
$C$ 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 the triangles $COA$ and $COB$.
3) Find the length of $CA$ and $CB$.
4) Find the area of triangle $ACB$
5) Compare the area of triangle $ACB$ found in part 2) to the sum of the areas found in part 2).

Solution
1)
Point $A$, $C$ and $B$ are on the circle of center $O$ and therefore $OA = OB = OC$. Hence $AOC$ and $COB$ are isosceles triangles and therefore
$\angle OCB = \angle CBO = 62^{\circ}$ which gives $\angle COB = 180 - 62 - 62 = 56^{\circ}$
$\angle COB$ and $\angle COA$ make a straight angle; hence $\angle COA = 180 - \angle COB = 124^{\circ}$
2)
Use the sine rule formula to find the area $A_1$ of triangle $COA$: $\quad A_1 = \dfrac{1}{2} \sin \angle COA \times OC \times OA \\ = \dfrac{1}{2} \sin 124^{\circ} \times 50 \times 50 = 1250 \sin 124^{\circ} \approx = 1036.30$ square units
In a similar way, the area $A_2$ of triangle $COB$ is given by: $\quad A_2 = \dfrac{1}{2} \sin \angle COB \times OC \times OB \\ = \dfrac{1}{2} \sin 56^{\circ} \times 50 \times 50 = 1250 \sin 56^{\circ} \approx 1036.30$ square units
3)
According to the semicircle theorem , $ACB$ is a right traingle with $AB$ as the hypotenuse. Hence $\sin 62^{\circ} = \dfrac{CA}{AB}$ and $\cos 62^{\circ} = \dfrac{CB}{AB}$
The above gives $CA = 100 \sin 62^{\circ}$ and $CB = 100 \cos 62^{\circ}$
4)
The area $A$ of the right triangle $ACB$ is given by: $A = \dfrac{1}{2} \times CA \times CB = \dfrac{1}{2} 100^2 \sin 62^{\circ} \cos 62^{\circ} \approx 2072.59$ square units
5)
The sum of the ares in part 2) is: $\quad A_1 + A_2 = 1036.30 + 1036.30 \approx 2072.60$ is very close to $A \approx 2072.59$. The small difference is due the rounding errors in the calculations.
In fact the sum of the areas $COA$ and $COB$ is equal to the area of triangle $ACB$.