Question

$\mathrm{CDE}$is an equilateral triangle formed on a side $\mathrm{CD}$ of a square $\mathrm{ABCD}$(Figure).

Show that $\mathrm{\Delta }\mathrm{ADE}\cong \mathrm{\Delta }\mathrm{BCE}$.

Answer 1

The congruency of Triangles:

Given:

$\mathrm{CDE}$ is an equilateral triangle formed on a side $\mathrm{CD}$ of a square $\mathrm{ABCD}$.

From $\mathrm{\Delta ADE}$ and $\mathrm{\Delta BCE}$,

• $\mathrm{DE}=\mathrm{CE}$ $\left[\mathrm{sides}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\right]$
  
• $$\begin{gathered} \angle \mathrm{ADC}=\angle \mathrm{BCD}=90° \\ \text{and}\angle \mathrm{EDC}=\angle \mathrm{ECD}=60° \\ \Rightarrow \angle \mathrm{ADC}+\angle \mathrm{EDC}=\angle \mathrm{BCD}+\angle \mathrm{ECD} \\ {150}^0={150}^0 \\ \therefore \angle \mathrm{ADE}=\angle \mathrm{BCE}\mathbf{[}\mathbf{Angle}\mathbf{}\mathbf{included}\mathbf{}\mathbf{between}\mathbf{}\mathbf{sides}\mathbf{]} \end{gathered}$$
  
• $\mathrm{AD}=\mathrm{BC}$ $\left[\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{square}\right]$

Therefore, $\mathrm{\Delta ADE}\cong \mathrm{\Delta BCE}$ $\left[\mathrm{by}\mathrm{SAS}\mathrm{rule}\right]$

Hence, it is proved that $\mathrm{\Delta ADE}\cong \mathrm{\Delta BCE}$