Saved Bookmarks
| 1. |
In Fig. 5, DEFG is a square and <BAC90°. Show that DE-BDEC.Fig. 5 |
|
Answer» Given: DEFG is a square and ∠BAC = 90°.To Prove: DE² = BD × EC.Proof : In ∆ AFG & ∆DBG ∠GAF = ∠BDG [ 90°]∠AGF = ∠DBG [corresponding angles because GF|| BC and AB is the transversal]∆AFG ~ ∆DBG [by AA Similarity Criterion] …………(1) In ∆ AGF & ∆EFC∠AFG = ∠CEF [ 90°]∠AFG = ∠ECF [corresponding angles because GF|| BC and AC is the transversal]∆AGF ~ ∆EFC [by AA Similarity Criterion] …………(2) From equation 1 and 2. ∆DBG ~ ∆EFCBD/EF = DG /ECBD/DE = DE /EC [ DEFG is a square] DE² = BD × EC Like my answer if you find it useful! |
|