$A B C$ and $D B C$ are two triangles in which $A C$ and $B D$ meet at $X$ . If $A X = D X$ and $B X = C X$ , then prove that $Δ A X B ≅ Δ D X C$ . $[2]$

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Solution

In $Δ B A X$ and $Δ C D X$ ,

$A X = D X \dots \dots \dots$ (given)

$B X = C X \dots \dots \dots$ (given)

$∠ A X B = ∠ D X C$ ……….(vertically opposite angles are equal) $[1 m a r k]$

$Δ B A X ≅ Δ C D X$ …… (SAS postulate) $[1 m a r k]$

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