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| 1. |
Prove that the ratio of the area of two similar triangle is equal to the square of its medians |
| Answer» Given: \u200b{tex}\\triangle {/tex}\u200bABC \u200b{tex} \\sim {/tex}\u200b {tex}\\triangle {/tex}DEFAM is a median in \u200b{tex}\\triangle {/tex}\u200bABC and DN is the corresponding median in {tex}\\triangle {/tex}DEFTo prove:{tex}\\frac{{area\\vartriangle ABC}}{{area\\vartriangle DEF}} = \\frac{{A{M^2}}}{{D{N^2}}}{/tex}Proof: \u200b{tex}\\triangle {/tex}\u200bDBC \u200b{tex} \\sim {/tex}\u200b {tex}\\triangle {/tex}DEF\u200b{tex}\\Rightarrow {/tex}\u200b\u200b{tex}\\angle{/tex}\u200b A = \u200b{tex}\\angle{/tex}\u200b D, \u200b{tex}\\angle{/tex}\u200b B = \u200b{tex}\\angle{/tex}\u200b E, \u200b{tex}\\angle{/tex}\u200b C = \u200b{tex}\\angle{/tex}\u200b F and{tex}\\frac{{AB}}{{DE}} = \\frac{{BC}}{{EF}} = \\frac{{AC}}{{DF}}{/tex}Also, {tex}\\frac{{area\\vartriangle ABC}}{{area\\vartriangle DEF}}{/tex}={tex}\\frac{{A{B^2}}}{{D{E^2}}} = \\frac{{B{C^2}}}{{E{F^2}}} = \\frac{{A{C^2}}}{{D{F^2}}}{/tex}........(i)[area theorem]Now,{tex}\\frac{{AB}}{{DE}} = \\frac{{BC}}{{EF}} = \\frac{{\\frac{1}{2}BC}}{{\\frac{1}{2}EF}} = \\frac{{BM}}{{EN}}{/tex}..............(ii)In {tex}\\triangle {/tex}ABM and DEN\u200b{tex}\\angle{/tex}\u200b B= \u200b{tex}\\angle{/tex}\u200b E and {tex}\\frac{{AB}}{{DE}} = \\frac{{BM}}{{EN}}{/tex} [From (ii)]\u200b{tex}\\Rightarrow {/tex}\u200b\u200b{tex}\\triangle {/tex}ABM \u200b{tex} \\sim {/tex}\u200b{tex}\\triangle {/tex}DEN [SAS similarity]\u200b{tex}\\Rightarrow {/tex}\u200b{tex}\\frac{{area\\vartriangle ABM}}{{area\\vartriangle DEN}} = \\frac{{A{B^2}}}{{D{E^2}}} = \\frac{{A{M^2}}}{{D{N^2}}} = \\frac{{B{M^2}}}{{E{N^2}}}{/tex}............(iii)From (i) and (iii), we get{tex}\\frac{{area\\vartriangle ABC}}{{area\\vartriangle DEF}} = \\frac{{A{M^2}}}{{D{N^2}}}{/tex} | |