From the figure shown, the line segments AB and CD intersect at point O. Angles AOD and COB are vertically opposite angles formed by the intersection of these two lines. Vertically opposite angles are always equal.

From the figure shown, the line segments AB and CD intersect at point O. Angles AOD and COB are vertically opposite angles formed by the intersection of these two lines. Vertically opposite angles are always equal.

Since AOD COB, their corresponding angles are equal. Angle A in AOD corresponds to angle C in COB, and angle D in AOD corresponds to angle B in COB. Thus, we conclude that A = C and B = D.

In Fig. 6.31, $OA \cdot OB = OC \cdot OD$ . Show that $\angle A = \angle C$ and $\angle B = \angle D$ . [Figure: segments AB and CD intersect at point O; A and D on one side, C and B on the other, forming triangles AOD and COB.]

Answer: It has been shown that $\angle A = \angle C$ and $\angle B = \angle D$ using the SAS similarity criterion for triangles $\triangle AOD$ and $\triangle COB$ .

Step-by-step solution

Step 1: Rearrange the given relation

We are given the product relationship $OA \cdot OB = OC \cdot OD$ . To use similarity criteria involving ratios of sides, we rearrange this equation by dividing both sides by $OC \cdot OB$ to form a proportion relating the sides.

Step 2: Identify vertically opposite angles

From the figure shown, the line segments AB and CD intersect at point O. Angles $\angle AOD$ and $\angle COB$ are vertically opposite angles formed by the intersection of these two lines. Vertically opposite angles are always equal.

Step 3: Apply SAS similarity criterion

We have established that $\frac{OA}{OC} = \frac{OD}{OB}$ (ratio of two sides are equal) and $\angle AOD = \angle COB$ (the included angles are equal). According to the SAS (Side-Angle-Side) similarity criterion, if two sides and the included angle of one triangle are proportional to the corresponding two sides and included angle of another triangle, then the two triangles are similar. Therefore, $\triangle AOD$ is similar to $\triangle COB$ .

Step 4: State corresponding angles of similar triangles

Since $\triangle AOD \sim \triangle COB$ , their corresponding angles are equal. Angle A in $\triangle AOD$ corresponds to angle C in $\triangle COB$ , and angle D in $\triangle AOD$ corresponds to angle B in $\triangle COB$ . Thus, we conclude that $\angle A = \angle C$ and $\angle B = \angle D$ .

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In Fig. 6.31, OA⋅OB=OC⋅ODOA \cdot OB = OC \cdot ODOA⋅OB=OC⋅OD. Show that ∠A=∠C\angle A = \angle C∠A=∠C and ∠B=∠D\angle B = \angle D∠B=∠D. [Figure: segments AB and CD intersect at point O; A and D on one side, C and B on the other, forming triangles AOD and COB.]