If in an equilateral triangle ABC, AD is median and in triangle ABD, AE is the median, then find the value of ABsq.: AEsq.

Answer :

Given ABC is a equilateral triangle , So median is also perpendicular on opposite side , So we form our diagram , As :

Here BD  =  DC  = $\frac{BC}{2} = \frac{AB}{2} = \frac{AC}{2}$ (  We know ABC is a equilateral triangle )

We apply Pythagoras theorem in triangle ABD and get 

AB²  =  BD² + AD²  

$$\begin{gathered} \Rightarrow A{B}^2 = {\left(\frac{AB}{2}\right)}^2 +\hspace{0.17em}A{D}^2 \\ \Rightarrow A{B}^2 = \frac{A{B}^2}{4} +\hspace{0.17em}A{D}^2 \\ \Rightarrow A{D}^2 = A{B}^2 - \frac{A{B}^2}{4} \\ \Rightarrow A{D}^2 = \frac{4 A{B}^{2 }- A{B}^2}{4} \\ \mathbf{\Rightarrow }\mathit{A}{\mathit{D}}^{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\frac{\mathbf{3}\mathbf{ }\mathbf{A}{\mathbf{B}}^{\mathbf{2}}}{\mathbf{4}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{)} \end{gathered}$$

And  In triangle AED 

BE  =  ED  =  $\frac{BD}{2} = \frac{AB}{4}$

We apply Pythagoras theorem in triangle AED and get 

AE²  =  ED² + AD² 

$$\begin{gathered} \Rightarrow {\mathrm{AE}}^2 = {\left(\frac{\mathrm{AB}}{4}\right)}^2 +\hspace{0.17em}{\mathrm{AD}}^2 \\ \Rightarrow {\mathrm{AE}}^2 = \frac{{\mathrm{AB}}^2}{16} +\hspace{0.17em}{\mathrm{AD}}^2 \\ \mathrm{Substitute} \mathrm{value} \mathrm{from} \mathrm{equation} 1 , \mathrm{and} \mathrm{get} \\ \Rightarrow {\mathrm{AE}}^2 = \frac{{\mathrm{AB}}^2}{16}+ \frac{{\mathrm{AB}}^2}{4} \\ \Rightarrow {\mathrm{AE}}^2 = \frac{{\mathrm{AB}}^{2 }+4 {\mathrm{AB}}^2}{16} \\ \Rightarrow {\mathrm{AE}}^2 = \frac{5 {\mathrm{AB}}^2}{16} \\ \Rightarrow \frac{{\mathrm{AB}}^2}{{\mathrm{AE}}^2} = \frac{16}{5} \\ \mathrm{So}, \\ {\mathbf{AB}}^{\mathbf{2}\mathbf{ }}\mathbf{:}\mathbf{ }{\mathbf{AE}}^{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{16}\mathbf{ }\mathbf{:}\mathbf{ }\mathbf{5}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)} \end{gathered}$$