, DEFG is a square and LRAC = 90°. Prove that :in figure(i) MGP-ADBG)

1.	, DEFG is a square and LRAC = 90°. Prove that :in figure(i) MGP-ADBG)ADBG AEFC(ii) AAGF AEFC(iv) DE- BD x EC.G900EC
Answer» 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.

, DEFG is a square and LRAC = 90°. Prove that :in figure(i) MGP-ADBG)ADBG AEFC(ii) AAGF AEFC(iv) DE- BD x EC.G900EC

Answer»

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.