InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Prove of BPT Theorem |
| Answer» Given :\xa0In {tex}\\triangle A B C{/tex}, DE || BC and intersects AB in D and AC in E.\xa0Prove that :\xa0{tex}\\frac{AD}{DB} = \\frac{AE}{EC}{/tex}Construction:\xa0Join BE, CD and draw EF {tex}\\perp{/tex} BA and DG {tex}\\perp{/tex} CA. Now from the given figure we have,EF {tex}\\perp{/tex} BA (Construction)EF is the height of ∆ADE and ∆DBE (Definition of perpendicular)Area({tex}\\triangle{/tex}ADE) ={tex}\\frac{AD.EF}{2}{/tex} .....(1)Area({tex}\\triangle{/tex}DBE) = {tex}\\frac{DB.EF}{2}{/tex} ....(2)Divide the two equations we have{tex}\\frac{Area \\triangle ADE}{Area \\triangle DBE} = \\frac{AD}{DB}{/tex} .....(3){tex}\\frac{Area \\triangle ADE}{Area \\triangle DEC} = \\frac{AE}{EC}{/tex} .....(4)Therefore, {tex}\\triangle \\mathrm{DBE} \\sim \\triangle \\mathrm{DEC}{/tex} (Both the ∆s are on the same base and\xa0between the same || lines).....(5)Area({tex}\\triangle{/tex}DBE) = Area({tex}\\triangle{/tex}DEC) (If the two triangles are similar their\xa0areas are equal){tex}\\frac{AD}{DB} = \\frac{AE}{EC}{/tex}\xa0[from equation 3,4 and 5]Hence proved. | |