Question

In the figure given above, if $AC=AD=CD=BD$; then $\angle ABC$ is ${30.5}^{\circ }$
If true then enter $1$ and if false then enter $0$

Answer 1

In $△$ ACD,
Since, AC = CD
$\angle CAD=\angle CDA=x$ (Angles opposite to equal sides are equal)
Sum of angles of the triangle,
$\angle ACD+\angle CAD+\angle CDA=180$
$58+2x=180$
$2x=122$
$x=61$
In $△ABD$,
AD = BD thus, $\angle ABD=\angle DAB=y$
Also, $\angle ABD+\angle DAB=\angle ADC$ (Exterior angle is equal to sum of opposite interior angles)
$2\angle ABD=61$
$\angle ABD={30.5}^{\circ }$