The perpendicular from A on side BC of triangle ABC intersect at D such that DB=CD .prove that 2AB

The perpendicular from A on side BC of triangle ABC intersect at D suc

1.	The perpendicular from A on side BC of triangle ABC intersect at D such that DB=CD .prove that 2AB
Answer» Given: The perpendicular from A on side BC of an{tex}\\vartriangle {/tex} ABC intersect BC at Such that BC = 3CDTo prove:2AB2 = 2AC2 + BC2\xa0Proof:In right triangle ADB,AB2 = AD2 + BC2 (1).......[By Pythagoras theorem]In right triangle ADC,AC2 = AD2 + CD2 (2).....[By Pythagoras theorem]Subtracting (2) from (1), we getAB2 - AC2 = BD2 - CD2= (BD + CD)(BD - CD)= (BC)(3CD - CD) ........ {tex}\\because {/tex} BD = 3CD(given)= (BC)(2CD) = 2(BC)(CD)=2(BC){tex}\\left( {\\frac{1}{4}BC} \\right){/tex}DB = 3CD{tex} \\Rightarrow \\frac{{BD}}{{CD}} = 3{/tex}{tex} \\Rightarrow \\frac{{DB}}{{CD}} + 1{/tex}= 3 + 1{tex} \\Rightarrow \\frac{{DB + CD}}{{CD}}=4{/tex}{tex}\\Rightarrow \\frac{{BC}}{{CD}} = 4{/tex}{tex}\\Rightarrow CD = \\frac{1}{4}BC = B{C^2}{/tex}{tex}\\Rightarrow {/tex} 2(AB2 - AC2) = BC2{tex}\\Rightarrow {/tex} 2AB2 - 2AC2 = BC2{tex}\\Rightarrow {/tex} 2AB2 = 2AC2 + BC2