Given: $¯ ¯¯¯¯¯¯ ¯ A C ≅ ¯ ¯¯¯¯¯¯¯ ¯ B D, ¯ ¯¯¯¯¯¯ ¯ B C ≅ ¯ ¯¯¯¯¯¯¯ ¯ A D$ . PRove that $¯ ¯¯¯¯¯¯¯ ¯ C K ≅ ¯ ¯¯¯¯¯¯¯ ¯ D K$

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Solution

given:- $A C = A D$
$B C = A C$
In $△ C A B$ $△ D A B$
$A B = A B$ (Common)
$A C = B D$ (given )
$B C = A D$ (given)
$∴ l y S S S$
$△ C A B ≅ △ D A B$
$l y C P C T \to$
$∠ A C B = ∠ A D B - - - - (1)$
In $△ A K C$ & $△ A B K C$
$A C = B D K (l y - - - (1))$
$∠ A K C = ∠ B K D$ (vertically opposite angle)
$∴ l y A A S -$
$△ A K C ≅ △ B K D$
$l y C P C T \Rightarrow$
$A K + B K$
$B C = A D$
$B C - A K = A D - A K$
$B C - B K = A D - A K (A K - B K)$
$C K - D K$

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