Prove that ratio of area of two similar triangle is equal to ratio of square of angle bisector

Prove that ratio of area of two similar triangle is equal to ratio of

1.	Prove that ratio of area of two similar triangle is equal to ratio of square of angle bisector
Answer» Given: {tex}\\triangle {/tex}ABC {tex} \\sim {/tex}{tex}\\triangle {/tex}DEFAM is the bisector of{tex}\\angle{/tex} BAC and is the corresponding bisector of {tex}\\angle{/tex} EDFTo prove:{tex}\\frac{{area\\vartriangle ABC}}{{area\\vartriangle DEF}} = \\frac{{A{M^2}}}{{D{N^2}}}{/tex}Proof: {tex}\\frac{{area\\vartriangle ABC}}{{area\\vartriangle DEF}} = \\frac{{A{B^2}}}{{D{E^2}}}{/tex}.......(i) [Area theorem]Now {tex}\\triangle {/tex}ABC {tex} \\sim {/tex}{tex}\\triangle {/tex}DEF{tex}\\Rightarrow {/tex}{tex}\\angle{/tex} A = {tex}\\angle{/tex} D {tex}\\Rightarrow {/tex}{tex}\\frac{1}{2}\\angle A = \\frac{1}{2}\\angle D{/tex}{tex}\\Rightarrow {/tex}{tex}\\angle{/tex}BAM = {tex}\\angle{/tex}EDNIn {tex}\\triangle {/tex}ABM and DEN {tex}\\angle{/tex} B = {tex}\\angle{/tex} E and {tex}\\angle{/tex} BAM= {tex}\\angle{/tex}EDN{tex}\\Rightarrow {/tex}{tex}\\triangle {/tex}ABM {tex} \\sim {/tex}{tex}\\triangle {/tex}DEN{tex}\\Rightarrow {/tex}{tex}\\frac{{AB}}{{DE}} = \\frac{{AM}}{{DN}} \\Rightarrow \\frac{{A{B^2}}}{{D{E^2}}} = \\frac{{A{M^2}}}{{D{N^2}}}{/tex}...................(ii){tex}\\Rightarrow {/tex}{tex}\\frac{{area\\vartriangle ABC}}{{area\\vartriangle DEF}} = \\frac{{A{M^2}}}{{D{N^2}}}{/tex} [From (i) and (ii)] Proved.