Question

In the diagram, ABD and BCD are isosceles triangles, where $AB=BC=BD$. What type of a quadrilateral is ABCD?
[2 Marks]

Answer 1

In, $\Delta BAD$
$\angle BDA=\angle BAD={57}^{\circ }$ [equal sides subtend equal angles]
In $\Delta BDC,$
$\angle BCD=\angle BDC={66}^{\circ }$ [equal sides subtend equal angles]
$\therefore \angle D={57}^{\circ }+{66}^{\circ }={123}^{\circ }$
$\angle A+\angle D={57}^{\circ }+{123}^{\circ }={180}^{\circ }$
Also, $\angle D+\angle C={123}^{\circ }+{66}^{\circ }={189}^{\circ }$
[1 Mark]

Hence, by the property that co-interior angles are supplementary, line AB$\parallel$DC,.
However, AD not parallel to BC.
Hence, ABCD is a trapezium.
[1 Mark]