$△$ ACD right angled at A and $△$ BDC right angled at B, and DA=CB= 4 c m ,CD= 8cm. Prove that $△$ ACD $≅ △$ BDC by RHS criterion. ( 2 Marks)

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Solution

In the two triangles, $△$ ACD and $△$ BDC,
side DA = side CB = 4cm
hypotenuse CD = hypotenuse DC = 8cm
$∠$ CAD = $∠$ CBD = 90° (1 Mark)

$∴$ $△$ ACD $≅ △$ BDC by RHS criterion for congruency of triangles
So, BD = AC (corresponding parts of congruent triangles.) (1 Mark)

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