In a ∆ ABC , suppose y = x is the equation of the bisector of the...

Given: Line $AC: 2x - y = 2 and BD: x = y$.

Line $AC\text{and}BD$ intersects at point $D$.

$⇒ 2x - x = 2$

$⇒ x = y = 2$

$⇒ D ≡ \left(2, 2\right)</p><p>We know that, by angle bisector theorem,$ the angle bisector of a triangle divides the base in the same ratio as the other two sides.

⇒ $\dfrac{CD}{AD} = \dfrac{BC}{AB}</p><p>It is given that,$2AB = BC$.</p><p>$⇒ $\dfrac{CD}{AD} = \dfrac{2AB}{AB}</p><p>$⇒ CD: AD = 2: 1$</p><p>Let the coordinates of$C$be \left(h, k\right).</p><p>By using section formula,</p><p>$D ≡ \left(\dfrac{h \times 1 + 2 \times 4}{3}, \dfrac{1 \times k + 2 \times 6}{3}\right) = \left(2, 2\right)

$⇒ h + 8 = 6, k + 12 = 6$

$⇒ h = - 2, k = - 6$

$⇒ C ≡ \left(- 2, - 6\right)$

Now, $2AB = BC$

$⇒ 2\sqrt{{\left(4 - \alpha\right)}^{2} + {\left(6 - \beta\right)}^{2}} = \sqrt{{\left(\alpha + 2\right)}^{2} + {\left(\beta + 6\right)}^{2}}$

Also, $B$ lines on $x = y$.

$⇒ 4\left[{\left(4 - \alpha\right)}^{2} + {\left(6 - \alpha\right)}^{2}\right] = {\left(\alpha + 2\right)}^{2} + {\left(\alpha + 6\right)}^{2}$

$⇒ 4\left(16 + {\alpha}^{2} - 8\alpha + 36 + {\alpha}^{2} - 12\alpha\right) = {\alpha}^{2} + 4 + 4\alpha + {\alpha}^{2} + 36 + 12\alpha$

$⇒ 8{\alpha}^{2} - 80\alpha + 208 = 2{\alpha}^{2} + 16\alpha + 40$

$⇒ 6{\alpha}^{2} - 96\alpha + 168 = 0$

$⇒ {\alpha}^{2} - 16\alpha + 28 = 0$

$⇒ \left(\alpha - 14\right)\left(\alpha - 2\right) = 0$

$⇒ \alpha = 2, 14$

But, coordinates of $B\text{and}D$ cannot be same.

$⇒ \alpha = \beta = 14$

$⇒ \alpha + 2\beta = 42$