Question

Angle BAC =90° of ∆ABC and DEFG is a square inside it prove that FG*FG=BG*FC

Answer 1

Answer:

Step-by-step explanation:

Our aim is to prove that FG^2 = BG*FC.

We know that ABC and that the angle ∠BAC is equal to 90°.

We are also given that DEFG is a square.

In order to prove that DE^2 = BD x EC, lets consider △AGF and △DBG.

We have the following:  

∠AGF = ∠GBD (corresponding angles)

∠GAF = ∠BDG (each = 90‘)

∴△AGF ~ △DBG .....(i)

Also, it is clear that △AFG ~ △ECF (AA Similarity).....(ii)

From (i) equation and (ii) equation, we can conclude that △DBG ~ △ECF.

BD/EF - BG/FC - DG/EC

BD/EF - DG/EC

EF × DG = BD × EC......(iii)

Similarly, since DEFG is a square, we get that:

DE = EF = FG = DG .....(iv)

From (iii) equation and equation (iv), DE^2 = BD × EC.  

Hence proved.

I hope this helps your studies!! Keep it up!

Answer 2

Answer:

FG*FG=BG*FC

Step-by-step explanation:

∠B = 90° - ∠C   or ∠C = 90° - ∠B

in  Δ BDG

ΔD = 90° - ∠B  = ∠C

& in Δ CEF

∠E = 90° - ∠C = ∠B

now comparing

Δ BDG   & Δ CEF

ΔD  = ∠C

∠B = ∠E

∠G = ∠F  = 90°

=> Δ BDG   ≈ Δ CEF

=> BD/CE  = BG/ EF  = DG/ FC

=> BG/ EF  = DG/ FC

=> BG * FC = EF * DG

EF = DG = FG  ( sides of square)

=> BG * FC = FG * FG

=> FG*FG=BG*FC

QED

proved