Question -
Answer -
Given, the perpendicular from A on side BC ofa┬а╬Ф ABC intersects BC at D such that;
DB = 3CD.
In ╬Ф ABC,
AD тКеBC and BD = 3CD
In right angle triangle, ADB and ADC, byPythagoras theorem,
AB2┬а=┬аAD2┬а+BD2┬атАжтАжтАжтАжтАжтАжтАжтАжтАж.(i)
AC2┬а=┬аAD2┬а+DC2┬атАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж..(ii)┬а
Subtracting equation┬а(ii)┬аfrom equation┬а(i), we get
AB2┬атАУ AC2┬а= BD2┬атАУDC2
┬а ┬а ┬а ┬а ┬а ┬а┬а ┬а ┬а ┬а= 9CD2┬атАУ CD2┬а[Since,BD = 3CD]┬а ┬а ┬а ┬а ┬а ┬а ┬а ┬а┬а
┬а ┬а ┬а ┬а ┬а ┬а ┬а ┬а ┬а ┬а= 8CD2┬а
= 8(BC/4)2┬а[Since, BC =DB┬а+ CD = 3CD┬а+ CD = 4CD]
Therefore, AB2┬атАУ AC2┬а= BC2/2
тЗТ 2(AB2┬атАУ AC2) = BC2
тЗТ 2AB2┬атАУ 2AC2┬а= BC2
тИ┤ 2AB2┬а= 2AC2┬а+ BC2.