Saved Bookmarks
| 1. |
25. In Figure, DEFG is a square and 4BAC 90° Show that DEBD x EC. |
|
Answer» Solution:-Given ; ABC is a triangle in which∠ BAC = 90° and DEFG is a square.Proof in : (1)InΔ AGF andΔ DBG,∠ AGF =∠ GBD (Corresponding angles)∠ GAF =∠ BDG = 90° eachSo,Δ AGF ~Δ DBG (ProvedBy AA similarity)(2)InΔ AGF andΔ EFC,∠ AFG =∠ FCE (Corresponding angles)∠ GAF =∠ CEF = 90° eachSo,Δ AGF ~Δ EFC (Proved by AA similarity)(3) InΔ DBG andΔ EFC,∠ DBG =∠ ECF = (Corresponding angles)∠ BDG =∠ CEF = 90° eachSo,Δ DBG ~Δ EFC (Proved by AA similarity)(4) InΔ AGF andΔ DBG,∠ AGF =∠ GBD (Corresponding angles)∠ GAF =∠ BDG = 90° each∴Δ AGF ~ΔDBG .....(1)Similarly,Δ AFG ~Δ ECF (AA similarity) ....(2)From (1) and (2), we getΔ DBG ~Δ ECF⇒BD/EF = BG/FC = DG/ECBD/EF = DG/ECEF× DG = BD× EC ....(3)Also DEFG is a square⇒ DE = EF = FG = DG ....(4)From (3)and(4), we getDE² = BD× ECHence proved. |
|