Question

Solve the question in attachment.​

Answer 1

Step-by-step explanation:

QUESTION :

• In the given fig the vertices of square

• DEFG are on the sides of AABC.< A = 90° then prove that DE² = BD x EC (Hint: Show that AGBD is similar to ACFE. Use GD = FE=DE).

given :

• DEFG are on the sides of AABC = 90 °

• DE² = BD x EC

to find :

• Show that AGBD is similar to ACFE. Use GD = FE=DE) = ?

• Show that AGBD is similar to ACFE. Use GD = FE=DE) = ?

solution :

• prove: DE2 = BD × EC

• <GDB = <GAF = 90 °

• <AGF = < GBF (corresponding,GF || BC

• and AB is a transversal line).

• By AA test of similarity

• AGBD ~AAGF

• CE/ DE = FE/BD

• CE/DE = DE/BD

• DE² = BD× CE

Answer 2

Question:-

In the adjoining figure, the vertices of square DEFG are on the sides △ ABC. ∠A = 90°. Then prove that DE² = BD × EC.(Hint:-Show that △GBD is similar to △CFE use GD = FE = DE).

Given:-

• A = 90°
• △ABC and DEFG is a square.

To Prove:-

• DE² = BD × EC.

Solution:-

Proof,

□ DEFG is a square.

GE || DE => Opposite sides of square are parallel.

i.e. GF || BC.

In △AGF & △DBG.

∠GAF = ∠BDF => (each of 90°).

∠AGF = ∠DBG => (Corresponding angle).

△AGF ~ △DBG => (By AA Test) => Equation (1).

In △AGC & △EFG.

∠GAF = ∠FEC => (each of 90°).

∠AFG = ∠ECF => (Corresponding angle).

△AGF ~ △EFC => (By AA Test). => Equation (2).

△DBG ~ △EFC => From Equation (1) & (2).

$\sf \frac{BD}{EF} = \frac{DG}{EC} \: = > corresponding \: sides \\ \: \sf \: of \: similiar \: triangles.$

EF × DG = BD × EC.

DE × DE = BD × EC => All sides of squares are congruent.

Answer:-

$\sf \fbox\red{Hence, Proved \: that \: DE = BD × EC.}$

Hope you have satisfied. ⚘