In the figure O is the midpoint of AB and CD. Prove that
(i) $△ A O C ≅ △ B O D$ ; (ii) $A C = B D$ .

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Solution

In triangles AOC and BOD, we have
AO = BO (O, the midpoint of AB);
$∠ A O C = ∠ B O D$ , (vertically opposite angles);
$C O = O D$ , (O, the midpoint of CD)
So by SAS postulate we have
$△ A O C ≅ △ B O D$ .
Hence, AC = BD, as they are corresponding parts of congruent triangles.

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In the figure O is the midpoint of AB and CD. Prove that(i) △AOC≅△BOD; (ii) AC=BD.

In triangles AOC and BOD, we haveAO = BO (O, the midpoint of AB);∠AOC=∠BOD, (vertically opposite angles);CO=OD, (O, the midpoint of CD)So by SAS postulate we have△AOC≅△BOD.Hence, AC = BD, as they are corresponding parts of congruent triangles.

In the figure O is the midpoint of AB and CD. Prove that
(i) $△ A O C ≅ △ B O D$ ; (ii) $A C = B D$ .

In triangles AOC and BOD, we have
AO = BO (O, the midpoint of AB);
$∠ A O C = ∠ B O D$ , (vertically opposite angles);
$C O = O D$ , (O, the midpoint of CD)
So by SAS postulate we have
$△ A O C ≅ △ B O D$ .
Hence, AC = BD, as they are corresponding parts of congruent triangles.