In the adjacent figure $△ A B C$ and $△ D B C$ are two triangles such that $¯ ¯¯¯¯¯¯ ¯ A B = ¯ ¯¯¯¯¯¯¯ ¯ B D$ and $¯ ¯¯¯¯¯¯ ¯ A C = ¯ ¯¯¯¯¯¯¯ ¯ C D$ . Show that $△ A B C ≅ △ D B C$

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Solution

Given ; $Δ A B C$ and $Δ D B C$ are two triangles such that $A B = B D$ and $A C = C D$ .
To show, $Δ A B C ≅ Δ D B C$
proof;
in triangle $D B M$ and triangle $A M C$ ,
$A B = B C$ (Given)
$A C = C D$ (Givne)
$B C = B C$ ( Common )
Therefore, by $S S S$ criterion of congruency, triangle $A B C$ is congruent to triangle $D B C$ .

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