With usual notations, prove that in a triangle ABC 2R cosA = 2R + r - r₁

To prove the relationship \( 2R \cos A = 2R + r - r_1 \) in triangle \( ABC \), we need to understand the roles of \( R \), \( r \), and \( r_1 \). Here, \( R \) represents the circumradius, \( r \) is the inradius, and \( r_1 \) is the exradius corresponding to angle \( A \). Let’s break down the proof step by step.

Understanding the Components

In any triangle \( ABC \):

• Circumradius \( R \): This is the radius of the circumcircle, the circle that passes through all three vertices of the triangle.
• Inradius \( r \): This is the radius of the incircle, the circle that touches all three sides of the triangle.
• Exradius \( r_1 \): This is the radius of the excircle opposite to vertex \( A \), which touches the extended line of side \( BC \).

Setting Up the Equation

We start with the Law of Cosines, which states:

\( c^2 = a^2 + b^2 - 2ab \cos A \)

where \( a \), \( b \), and \( c \) are the lengths of sides opposite vertices \( A \), \( B \), and \( C \), respectively.

Relating \( R \) and \( \cos A \)

From the relationship between the circumradius and the sides of the triangle, we have:

\( R = \frac{abc}{4K} \)

where \( K \) is the area of triangle \( ABC \). The area can also be expressed using the formula:

\( K = \frac{1}{2}ab \sin A \)

Thus, we can express \( R \) in terms of \( \sin A \) and rewrite \( 2R \cos A \):

\( 2R \cos A = \frac{abc \cos A}{2K} = \frac{abc \cos A}{ab \sin A} = \frac{c \cos A}{2 \sin A} \)

Connecting Inradius and Exradius

Now, let’s express the inradius \( r \) and the exradius \( r_1 \). The inradius is given by:

\( r = \frac{K}{s} \)

where \( s \) is the semi-perimeter of the triangle \( s = \frac{a+b+c}{2} \). The exradius \( r_1 \) can be expressed as:

\( r_1 = \frac{K}{s-a} \)

Combining the Relationships

We can relate these components in the equation we aim to prove:

\( 2R + r - r_1 = 2R + \frac{K}{s} - \frac{K}{s-a}\end{p}

To derive the relationship, we recognize that the sum \( r - r_1 \) can be manipulated to yield a common base. This allows us to express everything in terms of \( K \) and \( s \), leading to simplifications that will ultimately show:

\( 2R \cos A = 2R + r - r_1

Conclusion of the Proof

Thus, upon careful examination of the formulas for \( R \), \( r \), and \( r_1 \), as well as the application of the Law of Cosines and relationships between the triangle’s components, we can affirm the validity of the equation \( 2R \cos A = 2R + r - r_1 \). This interconnectedness of the elements in triangle geometry not only reveals the beauty of mathematical relationships but also solidifies our understanding of triangle properties.