B
A N
P
Given:
Quadrilateral inscribed in a circle,
AC perpendicular to BD,
Point M bisects DC.
D
M
C
Show: MP extended to N is perpendicular to AB.
Proof:
1. Inscribed angle ABD =angle ACD (labeled a) because they open on the same arc
AD.
2. Angle PAB, denoted a, is complementary to a because triangle APB is a right
triangle (Angle APB is a right angle, from the given AC is perpendicular to BD).
3. Inscribed angles BAC and BAC, denoted a, open on the same arc BC, and so are
equal.
4. Drop a perpendicular from point M to AC. Similarly, drop a
perpendicular from point M to BD. This will create a
quadrilateral with right angles at point q (by construction),
point r (by construction) and point p (given). Because the four
angles of a quadrilateral sum to 360º, the angle at point M is
also 90º, and the parallelogram is a rectangle.
5. Right triangle MDr is congruent to right triangle MCq by
angle-side-angle. We know the 2 angles other than the right
angle must be a and a, because a+a=90º. We have angle a,
side MD=CM (given), angle a.
6. Mr= Cq by "Corresponding Parts of Congruent Triangles"
7. The opposite sides of a rectangle are equal, so Mr = qp. But
from 6.) Mr= Cq, so Cq= qp (Things equal to the same thing
are equal to each other.)
8. Triangle Mqp is congruent to triangle MqC, by side-angle-side
(Mq=Mq, right angle q, Cq= qp)
9. Angle Mpq = angle MCq (denoted a), by CPCT
10. Angle ApN=angle Mpq (denoted a), by vertical angles.
11. The third angle ANp of triangle ANp is 90º because its other
two angles a and a add to 90º, and the sum of the three angles
add to 180º.
12. MN is perpendicular to AB, because it forms a right angle.