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Suppose that in \(\triangle A B C, \angle C=\angle A+\angle B\). Prove that \(C\) lies on the circle with diameter \(\underline{A B}\). (In particular, if the angles of \(\triangle A B C\) add to a straight angle, and \(\angle A C B\) is a right angle, then \(C\) lies on the circle with diameter \(\underline{A B}\).)

Short Answer

Expert verified
Since \(\angle ACB = 90^\circ\), \(C\) lies on the circle with diameter \(\underline{AB}\) by Thales' theorem.

Step by step solution

01

Understanding the Problem

We are asked to prove that point \(C\) lies on the circle with diameter \(\underline{AB}\) given that in the triangle \(\triangle ABC\), the angle \(\angle C\) is equal to the sum of \(\angle A\) and \(\angle B\).
02

Apply the Angle Sum Property

Recall that the sum of the angles in any triangle is \(180^\circ\), i.e., \( \angle A + \angle B + \angle C = 180^\circ \).
03

Substitute \(\angle C\)'s Expression

Since \(\angle C = \angle A + \angle B\), substitute this into the angle sum equation: \( \angle A + \angle B + (\angle A + \angle B) = 180^\circ \).
04

Simplify Equation

Simplifying, we have \(2(\angle A + \angle B) = 180^\circ\). So, \(\angle A + \angle B = 90^\circ\).
05

Understand Geometric Implication

If \(\angle A + \angle B = 90^\circ\), then \(\angle ACB\) will be a right angle (\(90^\circ\)).
06

Circle with Diameter Theorem

According to Thales' theorem, if \(C\) is a point such that \(\angle ACB = 90^\circ\), then \(C\) must lie on a circle with diameter \(\underline{AB}\).
07

Conclusion

Since \(\angle ACB = 90^\circ\), \(C\) lies on the semicircle with diameter \(\underline{AB}\). Hence, it is proven.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometry
Geometry is all about understanding the shapes, sizes, and properties of spaces and figures. It begins with points, lines, and planes, and extends to more complex shapes such as triangles, circles, and polygons. In the exercise about triangle \( \triangle ABC \), we use geometric principles to determine the relationship between the angles and the position of point \( C \) on a circle.
  • Geometric principles help us make sense of how different elements connect, like why some angles add up to 90 degrees, and how these angles relate to the position of points on circles.
  • Specifically, we learn that if the angles in a triangle lead us to identify a right angle, geometry can help us draw connections between this angle and the properties of a circle.
  • In this problem, geometric concepts have guided us to understand why a particular point in a triangle finds a place on a circle. It teaches us the elegant beauty of mathematics where different concepts intersect to provide a coherent result.
This exercise illustrates the power of geometry in linking the properties of shapes and guiding us to important conclusions through logical reasoning.
Triangle Properties
Triangles, one of the most fundamental shapes in geometry, have consistent properties that help us unlock many puzzles in math. A key trait of triangles is that the sum of its internal angles is always \(180^\circ\). This is crucial in our exercise, as it ties together the angles \( \angle A \), \( \angle B \), and \( \angle C \).

By understanding that \( \angle C = \angle A + \angle B \), and substituting into the triangles' angle sum equation, we derive that \( \angle A + \angle B = 90^\circ \). This conclusion implies that \( \angle ACB \) forms a right angle. Knowing the properties and limitations within triangles nuances our ability to solve such problems.
  • The exercise demonstrates how identifying relationships between angles can lead us to identify specific triangle types, such as right triangles.
  • Right triangles, in particular, often connect to circles through Thales’ theorem.
Thus, understanding triangle properties is not only about recognizing fundamental rules but also about identifying how these rules help solve problems involving different geometric shapes like circles.
Circle Theorems
A significant theorem in geometry is Thales' Theorem. It states that if \( C \) is a point on a circle such that \( \angle ACB = 90^\circ \), then \( C \) lies on a circle with its diameter being the line segment \( AB \). This is a classic example of how circle theorems help us connect linear and circular geometric principles.

In the exercise, once we determined that \( \angle ACB = 90^\circ \), Thales' theorem tells us confidently that point \( C \) must reside on a circle formed by the diameter \( AB \). This is a powerful insight that showcases the linking of triangle properties with circle behavior.
  • Circle theorems like Thales’ help dissolve the complexity by giving straightforward results for seemingly complex geometric shapes.
  • They provide proofs where we can visually and analytically see why certain points need to be where they are relative to circles.
Circle theorems are comprehensive, enabling us to explore relationships between straight lines and curves, enhancing our understanding of spatial aspects in geometry.

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Most popular questions from this chapter

Let \(\triangle A B C\) be an acute angled triangle. (a) Prove that, among all possible triangles \(\triangle P Q R\) inscribed in \(\triangle A B C,\) with \(P\) on \(B C, Q\) on \(C A, R\) on \(A B,\) the orthic triangle is the one with the shortest perimeter. (b) Suppose that the sides of \(\triangle A B C\) act like mirrors. A ray of light is shone along one side of the orthic triangle \(P Q,\) reflects off \(C A,\) and the reflected beam then reflects in turn off \(A B\). Where does the ray of light next hit the side \(B C ?\) (Alternatively, imagine the sides of the triangle as billiard table cushions, and explain the path followed by a ball which is projected, without spin, along \(P Q .)\)

Given any triangle \(\triangle A B C\) on the unit sphere with a right angle at the point \(A,\) we may position the sphere so that \(A\) lies on the equator, with \(\underline{A B}\) along the equator and \(\underline{A C}\) up a circle of longitude. Let \(O\) be the centre of the sphere and let \(\mathbf{T}\) be the tangent plane to the sphere at the point \(A\). Extend the radii \(\underline{O B}\) and \(\underline{O C}\) to meet the plane \(\mathbf{T}\) at \(B^{\prime}\) and \(C^{\prime}\) respectively. (a) Calculate the lengths of the line segments \(\underline{A B^{\prime}}\) and \(\underline{A C^{\prime}}\), and hence of \(\underline{B^{\prime} C^{\prime}}\) (b) Calculate the lengths of \(\underline{O B^{\prime}}\) and \(\underline{O C^{\prime}}\), and then apply the Cosine Rule to \(\triangle B^{\prime} O C^{\prime}\) to find an equation linking \(b\) and \(c\) with \(\angle B^{\prime} O C^{\prime}(=a) . \quad \)

(a) Find an expression involving "sin \(\frac{2 \pi}{n} "\) for the ratio \(\frac{\text { area of inscribed regular } n \text { -gon }}{\text { area of circumscribed circle }}\). (b) Find an expression involving "tan \(\frac{\pi}{n} "\) for the ratio \(\frac{\text { area of circumscribed regular } n \text { -gon }}{\text { area of inscribed circle }}\)

Given any triangle \(\triangle A B C,\) extend \(\underline{B C}\) beyond \(C\) to a point \(X\). Then the exterior angle $$\angle X C A=\angle A+\angle B$$ ("In any triangle, each exterior angle is equal to the sum of the two interior opposite angles.") Another important consequence is the result which underpins the sequence of "circle theorems".

Given points \(O, A,\) show how to construct the regular 3 -gon \(A B C\) with circumcentre \(O .\)

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