Question

In the diagram, $ABD$ and $BCD$ are isosceles triangles triangles, where $AB=BC=BD$. The special name that is given to quadrilateral $ABCD$ is:

• A. rectangle
• B. kite
• C. parallelogram
• D. trapezium

Answer 1

The correct option is D trapezium

In $\Delta BAD$,
$\angle BDA=\angle BAD={57}^{\circ }$ (isosceles triangle property)
In $\Delta BDC$,
$\angle BCD=\angle BDC={66}^{\circ }$ (isosceles triangle property)
$\therefore \angle D=\angle ADB+\angle BDC={57}^{\circ }+{66}^{\circ }={123}^{\circ }$ and
$\angle A+\angle D={57}^{\circ }+{123}^{\circ }={180}^{\circ }$
Also, $\angle D+\angle C={123}^{\circ }+{66}^{\circ }={189}^{\circ }$
Hence, by the property that corresponding interior angles are supplementry, lines are parallel, we have $AB\parallel DC$ and $AD\nparallel BC$
Hence, $ABCD$ is a trapezium.