Question

In the given, AB is diameter of the circle with centre O and CD || AB. If $\angle DAB={25}^o$, then find the measure of $\angle CAD.$

Answer 1

In $\Delta ADB$
we know that angle mathe on semicircle
${90}^o$ so
$\angle ADB={90}^o$
& $CD||AB$'
$\angle CDA=\angle DAB={25}^o$
$\angle CDA=90+25=115$
and ABCD is cyclic so
$\angle CAB+\angle CDB={180}^o$
$\angle CAD+\angle DAB+115=180$
$\angle CAD=180-115-25$
$\angle CAD={40}^o$