Answer:
There is a misprint in the question.
In the statement you have written DB=x ,and DA=y but in Question you have written DA= x and DB=y.
So, let me just considering your Statement justification
Let segment BC = a, segment CA = b ,segment AB = c segment CD = h, segment DB = x, segment AD = y ,y + x = c
In Δ B DC and Δ BC A
∠B D C =∠B C A [each being 90°]
∠ B is common.
Δ B D C is similar to Δ BC A.
[tex]\frac{BD}{BC}=\frac{BC}{BA}\\
\frac{x}{a}=\frac{a}{x}[/tex]
⇒ a² = c x .........(1)
Similarly we can prove that
Δ ADC is similar to Δ BC A.
⇒b²= c y ......(2)
adding (1) and (2)
⇒[tex]a^2+ b^2=cx+cy[/tex]
[tex]a^2+ b^2=c(x+y)[/tex]
=c×c [∴ x+y =c]
=c²
So, we have used two properties 1. Right Triangles Similarity Theorem 2.Substitution Addition Property of Equality.
we haven't used Side-Side-Side Similarity Theorem. It is not the right justification for the proof.
