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State the SAS criteria of similarity of two triangles. In the given figure, it is given that $OA \cdot OC = OB \cdot OD$. Use the SAS criteria to prove that $AD || CB$.
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Solution: Statement: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. [1 mark]
Given: $OA \cdot OC = OB \cdot OD$
To prove: $AD || CB$
Proof: In $\Delta OAD$ and $\Delta OBC$,
$OA \cdot OC = OB \cdot OD$ (given)
$\Rightarrow \frac{OA}{OB} = \frac{OD}{OC}$ [1 mark]
$\angle AOD = \angle BOC$ (vertically opposite angles) [1/2 mark]
$\Delta OAD \sim \Delta OBC$ (by SAS similarity criterion) [1 mark]
$\Rightarrow \angle A = \angle B$ [1 mark]
$AD || CB$ (alternate interior angles are equal) [1/2 mark]
Given: $OA \cdot OC = OB \cdot OD$
To prove: $AD || CB$
Proof: In $\Delta OAD$ and $\Delta OBC$,
$OA \cdot OC = OB \cdot OD$ (given)
$\Rightarrow \frac{OA}{OB} = \frac{OD}{OC}$ [1 mark]
$\angle AOD = \angle BOC$ (vertically opposite angles) [1/2 mark]
$\Delta OAD \sim \Delta OBC$ (by SAS similarity criterion) [1 mark]
$\Rightarrow \angle A = \angle B$ [1 mark]
$AD || CB$ (alternate interior angles are equal) [1/2 mark]