Question

In figure, $\square$ABCD is a parallelogram. $\angle ADB={40}^o,\angle ABD={30}^o$, find the measure of $\angle BAD-\angle DCB+\angle ADC-\angle ABC$

Answer 1

$\angle ADB+\angle DBA+\angle BAD={180}^o$ [ Sum of angles of triangle is ${180}^o$ ]

$\Rightarrow$ ${40}^o+{30}^o+\angle BAD={180}^o$

$\Rightarrow$ ${70}^o+\angle BAD={180}^o$

$\Rightarrow$ $\angle BAD={110}^o$

$\Rightarrow$ $\angle BAD=\angle DCB={110}^o$ [ Opposite angles of parallelogram are equal ]

$\Rightarrow$ $\angle BAD+\angle ADC={180}^o$ [ Adjacent angles of parallelogram ]

$\Rightarrow$ $\angle BAD+\angle ADB+\angle BDC={180}^o$

$\Rightarrow$ ${110}^o+{40}^o+\angle BDC={180}^o$

$\Rightarrow$ $\angle BDC={180}^o-{150}^o$

$\Rightarrow$ $\angle BDC={30}^o$

$\therefore$ $\angle ADC={40}^o+{30}^o={70}^o$

$\Rightarrow$ $\angle ADC=\angle ABC={70}^o$ [ Opposite angles of parallelogram are equal ]

$\therefore$ $\angle BAD={110}^o,\angle DCB={110}^o,\angle ADC={70}^o$ and $\angle ABC={70}^o$

$\therefore \angle BAD-\angle DCB+\angle ADC-\angle ABC=100-100+70-70=0$