In the given figure- the vertices of square DEFG are on the sides of - ABC - - A -90- - Then prove that DE 2- BD - EC Hint - Show that - GBD is similar to - CFE- Use GD - FE - DE-

Given: nbsp;▢DEFG is a square To prove: nbsp;DE2 nbsp;= BD × EC Proof: In nbsp;△GBD nbsp;and nbsp;△AGF ∠GDB = ∠GAF = nbsp; nbsp;90° nbsp; ...

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Standard X

Mathematics

Similar Triangles

In the given ...

Question

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

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Solution

Given: ▢DEFG is a square
To prove: DE² = BD × EC
Proof: In △GBD and △AGF
∠GDB = ∠GAF = 90° (Given)
∠AGF = ∠GBF (Corresponding, GF || BC and AB is a transversal line)
By AA test of similarity
△GBD ∼ △AGF ...(1)
In △CFEand △AGF
∠FEC = ∠GAF = 90° (Given)
∠FCE = ∠AGF (Corresponding, GF || BC and AC is a transversal line)
By AA test of similarity
△CFE ∼ △AGF ...(2)
From (1) and (2), we get
△CFE ∼ △GBD
$∴ \frac{CE}{GD} = \frac{FE}{BD} (Corresponding sides are proportional) \Rightarrow \frac{CE}{DE} = \frac{DE}{BD} (∵ GD = FE = DE) \Rightarrow {DE}^{2} = BD \times CE$
Hence proved.

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

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