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.

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Solution

$In △ ACD and △ FCB, we have : ∠ ACD = 90 ° + ∠ BCA and ∠ FCB = 90 ° + ∠ BCA$
∴ $∠ A C D = ∠ F C B$
$CF = CA (Sides of square) CD = CB (Sides of square) △ ACD ≅ △ FCB (SAS criterion)$
AD = BF (CPCT)
Hence, proved.

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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.

In △ACD and △FCB, we have:∠ACD=90°+∠BCAand ∠FCB=90°+∠BCA ∴ ∠ACD=∠FCB CF=CA (Sides of square)CD=CB (Sides of square)△ACD≅△FCB (SAS criterion) AD = BF (CPCT) Hence, proved.

$In △ ACD and △ FCB, we have : ∠ ACD = 90 ° + ∠ BCA and ∠ FCB = 90 ° + ∠ BCA$
∴ $∠ A C D = ∠ F C B$
$CF = CA (Sides of square) CD = CB (Sides of square) △ ACD ≅ △ FCB (SAS criterion)$
AD = BF (CPCT)
Hence, proved.