Suppose that the side lengths of a triangle are three consecutive...

Let $\text{B} = 2\text{A} and$\text{B}\text{D}$be the bisector of angle$\text{B}$, then<br /><br />$\text{CD} = \dfrac{\text{ab}}{\text{a} + \text{c}} \text{$and$} \text{AD} = \dfrac{\text{bc}}{\text{a} + \text{c}}$<br /><br />Now,$\Delta \text{ABC}$and \Delta$\text{BDC}$are similar, so<br /><br />$\dfrac{\text{BC}}{\text{AC}} = \dfrac{\text{CD}}{\text{BC}} ⇒ {\text{a}}^{2} = \dfrac{\text{ab}}{\text{a} + \text{c}} \text{b} ⇒ {\text{b}}^{2} = \text{a} \left(\text{a}$+$\text{c}\right) ......(i)

Since, \text{b} > \text{a} ⇒$Either$\text{b} = \text{a} +1$or$\text{b} = \text{a} +2,$if$\text{b} = \text{a} +$1,$ then [From Eq.(i)]

${\left(\text{a} +1\right)}$^{2} = \left(\text{a} + \text{c}\right) \text{a} ⇒ \text{c} =2 + \dfrac{1}{\text{a}}$<br /><br /> <img alt="" src="https://cdn-question-pool.Quantrex.app/nta_abhyas/jee_main/images/fvvvvvv.JPG" style="width: 300px; height: 169px;" /><br />$\text{c} i$s integer$⇒ \text{a} =1, \text{b} =2, \text{c} =3$but then, no triangle will form.<br />If$\text{b} = \text{a} +$2,$ then obviously \text{c} = \text{a} +1,$and then, [from Eq. (i)]<br /><br />${\left(\text{a}+2\right)}^{2} = \text{a}\left(2 $\text{a}$ +1\right)<br /><br />$⇒ {\text{a}}^{2} - 3a - 4 = 0\text{or}\text{a}=4$<br /><br />$∴\text{a}$$=4,$$\text{b}$$=6,$$\text{c}$$=5$ is the only possible solution.