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प्रश्न
In the figure, if ∠FEG ≡ ∠1 then, prove that DG2 = DE.DF
उत्तर
Proof:
| S.No. | Statements | Reasons |
| 1. |
∠FEG ≡ ∠1 ⇒ ∠DEG = 180° − ∠FE |
Given linear pair |
| 2. |
∠FDG + ∠DGF = 1 ∠EDG + ∠DFG = 1 |
Exterior angle = Sum of interior opposite angles ∵ ∠FDG = ∠EDG |
| 3. |
∠DEG = 180° − ∠FEG ∠DEG = 180° − ∠1 |
By 1 |
| 4. |
In ΔDFG ∠DEG = 180° − [∠FDG + ∠DFG] |
Angle sum property |
| 5. | ∠DGF = 180° − ∠1 | By 2 |
| 6. |
∠DGF = ∠DEG |
By 3 |
| 7. |
∠EDG = ∠EDG |
Common in ΔFDG and ΔEDG |
| 8. | ∴ ∠DGE = ∠DFG | Remaining angle by sum property and by 6 |
| 9. | ∴ ΔDGF ∼ ΔDEG |
By 6, 7, 8 By AAA similarity |
| 10. | `"DG"/"DE" = "GF"/"EG" = "DF"/"DG"` | Corresponding sides of similar traingle are proportional |
| 11. |
`"DG"/"DE" = "DF"/"DG"` DG.DG = DF.DE DG2 = DE.DF |
From 9 |
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