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.
You need to login to perform this action.
You will be redirected in
3 sec