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.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply

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

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

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

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

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

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

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

Log in to reply

zaffar u are super

Log in to reply

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

Log in to reply