Remember, while the questions may look complicated and it might be difficult to spot how to reach the answer, taking the time to apply these theorems will make your job a lot easier. When I learned these theorems, I never really understood where they came from – now you do! I think it’s remarkable what drawing a simple radius can unlock for us inside in a circle. In the larger triangle ∆BCD, we know that: This once again forms three isosceles triangles: ∆ABC, ∆ABD and ∆ACD. The proof starts in the same way, by drawing radii from the centre of the circle to each of the points B, C and D. The angle between the chord and the tangent is equal to the angle in the alternate segment:įinally, one of the more unexpected theorems we can derive from drawing lines in circles. We know that the sum of the interior angles of a quadrilateral is 360° (if you’re not sure about this, think of a square). This gives us four isosceles triangles: ∆ABC, ∆ACD, ∆ADE and ∆ABE. Opposite angles in a cyclic quadrilateral sum to 180 °:Īngle at B + angle at D = angle at C + angle at E = 180°Ī cyclic quadrilateral is a quadrilateral where all the corners are on the circumference of a circle.Īs before, the first step is to draw radii from the centre to each corner of the quadrilateral. Therefore, the angle at C is equal to the angle at D.Ĥ. We have that the angle at D is also half the angle at A. We can see that by applying the first theorem, we have that the angle at C is half the angle at A (the centre). The way we show this is by using the first theorem. If we add equations (1) and (2), we have b+2s+a+2t = 360°. (1)Īnd for ∆ACD we have that a+2t=180° (180° in a triangle). įor ∆ABC, we can see that b+2s=180° (180° in a triangle). We also have that ∆ABC and ∆ACD are isosceles.įirstly, we see that a+b=180° (180° on a straight line). This makes three triangles: ∆ABC, ∆ACD and a large one, ∆BCD. Note that this is a radius of the circle. If we wanted to show this without using Theorem 1, start by drawing a line from A to C. The angle in a semicircle is a right angle:įirstly, we can see that this is an application of the theorem above, with angle at the centre = 180°. Īnd we have in the larger triangle ∆DCB that: So we can see from ∆ABC that the angle at A is 180-2s (180° in a triangle). Since the lines AB, AC and AD are all radii of the circle, this means that the triangles ∆ACD, ∆ABD and ∆ABC are isosceles. This forms three small triangles (∆ACD, ∆ABC, ∆ABD) and one big one (∆DCB). How do we show this? Start by drawing lines to connect A and D, and B and C. The angle at the centre is twice the angle at the circumference: The angle between the chord and the tangent is equal to the angle in the alternate segmentġ. Opposite angles in a cyclic quadrilateral sum to 180° 5.
The angle in a semicircle is a right angle 3. The angle at the centre is twice the angle at the circumference 2. Since every radius is the same, drawing two radii forms a triangle with two equal sides – an isosceles triangle! We’ll be doing this a lot, so here’s an example: Ready? Let’s go.įirstly, we have to know how to construct an isosceles triangle from two radii. The defining feature of the circle is its constant radius, and I hope to show you that starting from this simple line, we can derive all the circle theorems you need to understand. Once we draw some lines inside a circle, we can deduce patterns and theorems that are useful both theoretically and in a practical sense. It’s so simple to understand, but it also gives us one of the most crucial constants in all of mathematics: p. In my opinion, the most important shape in maths is the circle.
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