Saved Bookmarks
| 1. |
In figure angleBAC=90degree show DE square=BD intoEC |
| Answer» Given: {tex}\\triangle {/tex}ABC is right-angled at A and DEFG is a squareTo Prove: DE2 = BD {tex} \\times{/tex} ECProof: Let {tex}\\angle{/tex} C = x ........(i)Then, {tex}\\angle{/tex} ABC = 90o - x[ {tex}\\because {/tex} ABC is right angled]Also {tex}\\triangle {/tex} BDG is right-angled at D.{tex}\\angle{/tex}BGD = 90o - (90o - x) = x ..............(ii)From (i) and (ii), we get{tex}\\angle{/tex}BGD = {tex}\\angle{/tex}C.............(iii)Consider {tex}\\triangle {/tex} BGD and {tex}\\triangle {/tex} CEF{tex}\\angle{/tex}CEF = {tex}\\angle{/tex}BDG = 90o [ {tex}\\because {/tex}DEFG is square]{tex}\\angle{/tex}BGD = {tex}\\angle{/tex}C [From (iii)]{tex}\\therefore {/tex}{tex}\\triangle {/tex}BGD {tex}\\cong{/tex}{tex}\\triangle {/tex}FEC [By AA similarity]{tex}\\therefore {/tex}{tex}\\frac{{BD}}{{EF}} = \\frac{{DG}}{{EC}}{/tex}{tex}\\Rightarrow {/tex} EF{tex} \\times{/tex} DG = BD {tex} \\times{/tex} ECBut EF = DG {tex} \\times{/tex} DE [ {tex}\\because {/tex} side of a square]{tex}\\Rightarrow {/tex} DE {tex} \\times{/tex} DE = BD {tex} \\times{/tex} EC{tex}\\Rightarrow {/tex} DE2 = BD {tex} \\times{/tex} EC | |