$\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$ in a triangle

Find the angle $\alpha $ of a triangle with sides a, b and c for which the equality

Find the angle $\alpha $ of a triangle with sides a, b and c for which the equality

Sam Hardin
2022-07-10
Answered

$\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$ in a triangle

Find the angle $\alpha $ of a triangle with sides a, b and c for which the equality

Find the angle $\alpha $ of a triangle with sides a, b and c for which the equality

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Remark 2: Euclid's proof amounts to solving the above equation. Construct the square ABCD, given segment AB. Let E be the midpoint of AD. Construct the point of intersection, F, of the dircde centered at E of radius EB and the extension of segments DA, as shown. Now construct square AFGX, all of whose sides have length AF, with X on segment AB.

Claim: X is the desired point.

Notice that $AE=\frac{1}{2}AD=\frac{1}{2}a$

(ii)

Apply the Pythagorean Theorem to the right triangle ABE to find EB in terms of a.

EB=______

Claim: X is the desired point.

Notice that $AE=\frac{1}{2}AD=\frac{1}{2}a$

(ii)

Apply the Pythagorean Theorem to the right triangle ABE to find EB in terms of a.

EB=______

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