Question

In the given figure, DEFG is a square and BAC = 900 . Show that FG2 = BG x Fc

Answer 1

Hi there,

The figure required for the question is missing. I have attached a figure below that satisfies the question given and have solved it accordingly.

Step-by-step explanation:

Given data:

DEFG is a square i.e., DE = EF = FG = GD &  ∠GDE = ∠DEF = ∠EFG = ∠FGD = 90° ….. [∵ all sides of a square are equal in length and angles are equal to 90°]

∠BAC = 90°

To show: FG² = BG x FC

Solution:

Step 1:

In ∆AGF and ∆GDB,

∠A = ∠GDB = 90°

∠AGF = ∠GBD ….. [corresponding angles]

∴ By AA similarity, ∆AGF ~ ∆GDB ……. (i)

Step 2:

In ∆AGF and ∆FCE,

∠A = ∠FEC = 90°

∠AFG = ∠FCE ….. [corresponding angles]

∴ By AA similarity, ∆AGF ~ ∆FCE ……. (ii)

Step 3:

From (i) & (ii), we get

∆GDB ~ ∆FCE

Since corresponding sides of two similar triangles are proportional

∴ GD/FC = BG/EF

⇒ GD * EF = BG * FC

⇒ FG² = BG * FC …… [∵ GD = EF = FG]

Hope this is helpful!!!!!

Answer 2

Answer:

Hope this helps

Step-by-step explanation: