question_answer

In an equilateral \[\Delta \text{ }ABC,\text{ }E\] is any point on \[BC\] such that \[BE=\frac{1}{4}BC\]. Prove that \[16\text{ }A{{E}^{2}}=13\text{ }A{{B}^{2}}\].

Answer:

Given          \[BE=\frac{1}{4}BC\]

Draw           \[AD\bot BC\]

In \[\Delta \text{ }AED\] by pythagoras theorem,

\[A{{E}^{2}}=A{{D}^{2}}+D{{E}^{2}}\]                                                                         ...(i)

In \[\Delta \,ADB\]

\[A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}\]

\[A{{B}^{2}}=A{{E}^{2}}-D{{E}^{2}}+B{{D}^{2}}\]         [From (i)]

\[=A{{E}^{2}}-D{{E}^{2}}+{{(BE+DE)}^{2}}\]

\[A{{B}^{2}}=A{{E}^{2}}-D{{E}^{2}}+B{{E}^{2}}+D{{E}^{2}}+2BE.DE\]

\[A{{B}^{2}}=A{{E}^{2}}+B{{E}^{2}}+2BE.DE\]

\[A{{B}^{2}}=A{{E}^{2}}+{{\left( \frac{BC}{4} \right)}^{2}}+2\frac{BC}{2}(BD-BE)\]

\[A{{B}^{2}}=A{{E}^{2}}\frac{B{{C}^{2}}}{16}+\frac{BC}{2}\left( \frac{BC}{2}-\frac{BC}{4} \right)\]

\[A{{B}^{2}}=A{{E}^{2}}+\frac{A{{B}^{2}}}{16}+\frac{AB}{2}\left[ \frac{2\,\,AB-AB}{4} \right]\]

\[A{{B}^{2}}=A{{E}^{2}}+\frac{AB}{16}+\frac{AB}{2}\times \frac{AB}{4}\]

\[A{{B}^{2}}+\frac{A{{B}^{2}}}{16}+\frac{A{{B}^{2}}}{8}=A{{E}^{2}}\]

\[\frac{16\,\,A{{B}^{2}}-A{{B}^{2}}-2A{{B}^{2}}}{16}=A{{E}^{2}}\]

\[16\,\,A{{B}^{2}}-3\,A{{B}^{2}}=16\,\,A{{E}^{2}}\]

\[13\,\,A{{B}^{2}}=16\,\,A{{E}^{2}}\]                       Hence Proved.