Question

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

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

Answer 1

Answer: (C)

$AB=BC+AD$

Extend the sides $AD$ and $BC$ to complete the equilateral triangle $△PAB$.

Let the side $AB$ be $x$, $AD$ be $a$ and $BC$ be $b$.

Observe that,$△ABC\sim △BPD$

i.e., $\frac{BC}{PD}=\frac{AB}{BP}\Rightarrow \frac{b}{x-a}=\frac{x}{x}\Rightarrow x=a+b$

Therefore $AB=BC+AD$