a We are given that in ∘ C, radii CA and CB are perpendicular. Let's draw a circle and choose the point A that lies on this circle. Segment CA is a radius of ∘ C.
To find point B we can use the set square, keeping in mind that we want radii CA and CB to be perpendicular.
Next we know that BD and AD are tangent of ∘ C. Let's recall that a tangent line is perpendicular to a radius of the circle. Therefore, we can again use the set square to draw lines that are perpendicular to CA and CB.
The point of intersection of the dashed lines will be point D.
b In this part we want to determine what type of quadrilateral CABD is. First, let's recall that ∠ D also needs to be right, as the sum of the angle measures in any quadrilateral is equal to 360^(∘).
As we can see in this quadrilateral, opposite angles are congruent so this quadrilateral is a parallelogram. Additionally, all angles in this parallelogram are right, so ABCD is a rectangle. Next we know that AC=CB, as these segments are radii of ∘ C.
By the External Tangent Congruence Theorem we know that AD and BD are congruent. However, since in a rectangle opposite sides are congruent, this indicates that AD≅ BD≅ CB≅ AC.
Finally, as all sides in a rectangle CADB are congruent, this quadrilateral is a square.
