Mathisconfusing101 said:
what about this one?
In the figure provided, DC and BC are tangents of circle A and DB joins the points of tangency. Classify triangle CEB.
Since I don't know which theorems you may already have covered, I'll go through it the "long way":
AD and AB are congruent because of the definition of a a circle (all points on a circle are the same distance from the center)
/ ADC and
/ ABC are right angles because a radius drawn to the point of tangency is perpendicular to the tangent, and perpendicular lines form right angles.
AC is congruent to AC by the reflexive property of congruence (any segment is congruent to itself).
Triangle ADC is congruent to triangle ABC by the Hypotenuse-Leg congruence theorem.
DC is congruent to BC and
/ DCE is congruent to
/ BCE because corresponding parts of congruent triangles are congruent.
Now, look at triangles DCE and BCE. You already know that we have one pair of congruent angles (the angles at C) and one pair of congruent sides, DC and BC. And, EC is congruent to EC. So, triangles DCE and BCE are congruent by the Side-Angle-Side congruence postulate (or theorem, depending on your book).
/ DEC is congruent to
/ BEC because they are corresponding parts of congruent triangles. They are also supplementary, because they form a linear pair. If two angles are congruent and supplementary, then each is a right angle; so triangle CEB is a RIGHT triangle.
And, jonboy, HUH??
