Question

In the above figure, O is the center of the circle, $\overline{AB}$ is tangent to the circle at D, and $\overline{BC}$ is tangent to the circle at E. Which of the following must be true?

• A. $\overline{OC}\equiv \overline{OD}$
• B. $△ABC$ is an isosceles triangle
• C. $△ABC\equiv △OEC$
• D. $△OEC\equiv △ADO$
• E. $\square DBEO$ is a square

Answer 1

The correct option is E $\square DBEO$ is a square

WKT

$\angle OEB=\angle OEC=\angle ODA=\angle ODB=90$

Because radius and the tangent are perpendicular to each other at the point where tangent touches the circle

And given $\angle EDB=90$

All the angles in BEDO are right angles

And $OE=OD=radius$

Adjacent sides are also equal

$DBEO$ is a square.