Question

*. In
In figure1.81, the vertices of
square DEFG are on the sides of
A ABC. ZA = 90°. Then prove that
DE? = BD X EC
(Hint : Show that A GBD is similar
to A CFE. Use GD = FE = DE.)​

Answer 1

Answer:

Given: ABC is a triangle in which ∠BAC = 90° and DEFG is a square.

To prove: DE2 = BD x EC.

Proof: In △AGF and △DBG,

∠AGF = ∠GBD (corresponding angles)

∠GAF = ∠BDG (each = 90‘)

∴△AGF ~ △DBG .....(i)

Similarly, △AFG ~ △ECF (AA Similarity).....(ii)

From (i) and (ii), △DBG ~ △ECF.

BD/EF - BG/FC - DG/EC

BD/EF - DG/EC

EF × DG = BD × EC......(iii)

Also DEFG is a square

⇒ DE = EF = FG = DG .....(iv)

From (iii) and (iv),

DE2 = BD × EC. _____________proved

Step-by-step explanation:

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