2002 Pan African MO Problems/Problem 5
Problem
Let be an acute angled triangle. The circle with diameter AB intersects the sides AC and BC at points E and F respectively. The tangents drawn to the circle through E and F intersect at P. Show that P lies on the altitude through the vertex C.
Solution
Draw lines and , where and are on and , respectively. Because and are tangents as well as and , and . Additionally, because and are tangents, .
Let and . By the Base Angle Theorem, and . Additionally, from the property of tangent lines, , , , and . Thus, by the Angle Addition Postulate, and . Thus, and , so . Since the sum of the angles in a quadrilateral is 360 degrees, . Additionally, by the Vertical Angle Theorem, and . Thus, .
Now we need to prove that is the center of a circle that passes through . Extend line , and draw point not on such that is on the circle with . By the Triangle Angle Sum Theorem and Base Angle Theorem, . Additionally, note that , and since , . Thus, by the Base Angle Converse, . Furthermore, . Therefore, is the diameter of the circle, making the radius of the circle. Since is a point on the circle, .
Thus, by the Base Angle Theorem, , so . Since , by the Alternating Interior Angle Converse, . Therefore, since , , and must be on the altitude of that is through vertex .
See Also
2002 Pan African MO (Problems) | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All Pan African MO Problems and Solutions |