In the given figure, ABC is a triangle, right angled at B. If BCDE is a square on side BC and ACFG is a square on AC, prove that AD=BF.
Given: ∠ABC = 90° , BCDE is a square on side BC and ACFG is a square on AC
To prove: AD = EF
Proof:
Since BCDE is square,
∠BCD = 90° …(1)
In ∆ACD,
∠ACD = ∠ACB + ∠BCD
= ∠ACB + 90° …(2)
In ∆BCF,
∠BCF = ∠BCA + ∠ACF
Since ACFG is square,
∠ACF = 90° …(3)
From 2 and 3, we have,
∠ACD = ∠BCF ….(4)
Thus in ∆ACD and ∆BCF, we have,
AC = CF ...sides of square
∠ACD = ∠BCF …from 4
CD = BC … sides of square
Thus by SAS property of congruence,
∆ACD ≅ ∆BCF
Hence, we know that, corresponding parts of the congruent triangles are equal
∴ AD = BF