In a triangle \( ABC \) , \( \angle A \) is twice of \( \angle B \) and \( a, b \) and \( c \) are respective sides opposite to angles \( \angle A, \angle B \) and \( \angle C \). then prove that \( a^2 = b(b+c) \)

Let D be on BC such that AD bisects \(\angle A\). Then \(\triangle ABD\) is isoceles. Also, \(\triangle ADC\) is similar to \(\triangle BAC\) as all their corresponding interior angles are equal.

By angle bisector theorem CD=\(\frac {ab}{b+c}\). However, since \(\triangle ACD\) is similar to \(\triangle BCA\), \(\frac {a}{b}=\frac {b}{\frac {ab}{b+c}}=\frac {b+c}{a}\). Cross multiplying gives the result.

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TopNewestLet D be on BC such that AD bisects \(\angle A\). Then \(\triangle ABD\) is isoceles. Also, \(\triangle ADC\) is similar to \(\triangle BAC\) as all their corresponding interior angles are equal.

By angle bisector theorem CD=\(\frac {ab}{b+c}\). However, since \(\triangle ACD\) is similar to \(\triangle BCA\), \(\frac {a}{b}=\frac {b}{\frac {ab}{b+c}}=\frac {b+c}{a}\). Cross multiplying gives the result.

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One of the proofs is this.. Does anyone have a simpler proof?

Let \( \angle A = 2x \) , \( \angle B = x \)

From Sine rule,

\( \frac{Sin A}{a} = \frac{Sin B}{b} \)

\( \implies \frac{Sin A}{Sin B} = \frac{a}{b} \)

\( \implies \frac{Sin 2x}{Sin x} = \frac{a}{b} \)

\( \implies \frac{2 Cos x Sin x}{Sin x} = \frac{a}{b} \)

\( \implies 2 Cos x = \frac{a}{b} \) ...(1)

From Cosine rule,

\( Cos A = \frac{b^2 + c^2 - a^2}{2bc} \) ...(2)

\( Cos B = Cos x = \frac{a^2 + c^2 - b^2}{2ac} \) ...(3)

Now, \( Cos A = Cos 2x = Cos^{2} x - Sin^{2} x \)

\( \implies (Cos 2x) + 1 \)

\( = [ Cos^{2} x - Sin^{2} ] + [ Cos^{2} x + Sin^{2} x ] = 2 Cos^{2} x \)

From (1), \( 2 Cos^{2} x = ( Cos 2x ) + 1 = \frac{b^2 + c^2 - a^2}{2bc} + 1 \)

\( = 2 Cos^{2} x = \frac{b^2 + c^2 - a^2 + 2bc}{2bc} \) ...(4)

Dividing (4) with (3),

\( \frac{2 Cos^{2} x}{Cos x} \)

\( = 2 Cos x = \frac{b^2 + c^2 - a^2 + 2bc}{a^2 + c^2 - b^2} \times \frac{2ac}{2bc} \)

\( = 2 Cos x = \frac{b^2 + c^2 - a^2 + 2bc}{a^2 + c^2 - b^2} \times \frac{a}{b} \)

From (1), \( 2 Cos x = \frac{a}{b} \)

\( \implies \frac{a}{b} = \frac{b^2 + c^2 - a^2 + 2bc}{a^2 + c^2 - b^2} \times \frac{a}{b} \)

\( \implies \frac{b^2 + c^2 - a^2 + 2bc}{a^2 + c^2 - b^2} = 1 \)

\( \implies a^2 + c^2 - b^2 = b^2 + c^2 - a^2 + 2bc \)

\( \implies 2a^2 = 2b^2 + 2bc \)

\( \implies a^2 = b^2 + bc \)

\( = \boxed{a^2 = b(b+c)} \)

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zaffar u are super

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Nice use of trigonometry, although similar triangles would quicken the process

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