### BBC Bitesize - GCSE Maths - Circle theorems - Higher - Edexcel - Revision 5

Proof showing that a triangle inscribed in a circle having a diameter as one side Inscribed shapes: angle subtended by diameter Just lacquer and semi- precious stones, as indicated in the credits at . Well, we have in our tool kit the notion of an inscribed angle, it's relation to a central angle that subtends the same arc. Explaining circle theorem including tangents, sectors, angles and proofs, with Two Radii and a chord make an isosceles triangle. Angle in a Semi-Circle from the ends of the diameter of a circle to its circumference form a right angle. The angle inscribed in a semicircle is always a right angle (90°). Try this Drag any orange The triangle formed by the diameter and the inscribed angle ( triangle ABC above) is always a right triangle. Relationship to Thales' Theorem. This is a.

This common endpoint forms the vertex of the inscribed angle.

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The other two endpoints define what we call an intercepted arc on the circle. The intercepted arc might be thought of as the part of the circle which is "inside" the inscribed angle. See the pink part of the circle in the picture above. A central angle is any angle whose vertex is located at the center of a circle. A central angle necessarily passes through two points on the circle, which in turn divide the circle into two arcs: The minor arc is the smaller of the two arcs, while the major arc is the bigger.

We define the arc angle to be the measure of the central angle which intercepts it. That and that must be the same, or if I were to draw it up here, that and that must be the exact same base angle.

Now let me see, I already used theta, maybe I'll use x for these angles.

So this has to be x, and that has to be x. So what is x going to be equal to? Well, x plus x plus 2theta have to equal degrees. They're all in the same triangle.

So let me write that down.

### Circle Theorems - Mathematics GCSE Revision

We get x plus x plus 2theta, all have to be equal to degrees, or we get 2x plus 2theta is equal to degrees, or we get 2x is equal to minus 2theta. Divide both sides by 2, you get x is equal to 90 minus theta.

So x is equal to 90 minus theta. Now let's see what else we could do with this. Well we could look at this triangle right here.

This triangle, this side over here also has this distance right here is also a radius of the circle. This distance over here we've already labeled it, is a radius of a circle. So once again, this is also an isosceles triangle. These two sides are equal, so these two base angles have to be equal.

### Proof that the angle of a triangle in a semicircle is 90°

So if this is theta, this is also going to be equal to theta. And actually, we use that information, we use to actually show that first result about inscribed angles and the relation between them and central angles subtending the same arc.

So if this is theta, that's theta because this is an isosceles triangle. So what is this whole angle over here?

## Proof that the angle of a triangle in a semicircle is 90°.

Proof of the converse using geometry[ edit ] Figure for the proof of the converse This proof consists of 'completing' the right triangle to form a rectangle and noticing that the center of that rectangle is equidistant from the vertices and so is the center of the circumscribing circle of the original triangle, it utilizes two facts: Let D be the point of intersection of lines r and s Note that it has not been proven that D lies on the circle The quadrilateral ABCD forms a parallelogram by construction as opposite sides are parallel.

Then the point O, by the second fact above, is equidistant from A,B, and C. And so O is center of the circumscribing circle, and the hypotenuse of the triangle AC is a diameter of the circle. Let O be the center of C.

Let D be the intersection of C and the ray OB. But then D must equal B. Proof of the converse using linear algebra[ edit ] This proof utilizes two facts: Let M's center lie on the origin, for easier calculation. This means that A and B are equidistant from the origin, i.

Since A lies on M, so does B, and the circle M is therefore the triangle's circumcircle.