Question

The diagonals of a cyclic quadrilateral ABCD are congruent. Show that AD=BC and segment AB is parallel to segment CD

Answer 1

For a cyclic quadrilateral ABCD we have shown that AD=BC and the segment AB║CD is proved

Explanation:

Given that the diagonals of a cyclic quadrilateral ABCD are congruent To prove that AD=BC and the segment AB║CD

• From the figure we have AC and BD are the diagonals of the given quadrilateral ABCD Hence we have AC=BD and AD=BC ( given )

• Therefore in the quadrilateral the triangles are ADC and BCD we have CD=CD

Therefore the the triangles ADC≅BCD ( by SSS criteria " two triangles are congruent if all corresponding sides are equal" )

Therefore  ∠D=∠C ( by SAS criteria " two triangles are congruent if two corresponding sides and the angle in between are equal" )

• Given that ABCD is a cyclic quadrilateral  and  ∠C=∠D Now by the property of cyclic quadrilateral "The sum of the opposite angles of a cyclic quadrilateral is supplementary".

• We have  ∠C+∠D=180°
• ∠C+∠C=180° ( since ∠C=∠D )

• 2∠C=180°

• $C=\frac{180^{\circ}}{2}$

• ∴ ∠C=90°
• ∠C=∠D=90° But they are interior angles with 90°

Hence the segment AB║CD is proved.

Answer 2

Answer:here is it

Explanation:

The best and perfect answer

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