Question

In a quadrilateral ABCD, which is not a trapezium, it is known that $\angle$ DAB = $\angle$ABC = ${60}^0$. Moreover, $\angle$CAB = $\angle$CBD. Then,

• A. AB = BC + CD
• B. AB = AD + CD
• C. AB = BC +AD
• D. AB = AC + AD

Answer 1

The correct option is D

AB = AC + AD

Extend the sides $AD$ and $BC$ to complete the equilateral triangle $△AB{D}^{\prime }$.
Now from above figure $△BAC\cong △D^{\prime }BD$
$\therefore D{D}^{\prime }=BC\Rightarrow A{D}^{\prime }-AD=BC\Rightarrow AB=AD+BC$