Two circles meet at A and B. A chord CD of one circle is produced to meet the other circle at E and F so that CDEF is a straight line, as shown. The common chord AB is produced to meet the line CF at a point M between D and E. If M is the midpoint of CF and angle CAF=90 degrees, prove that AC//BD and AF//BE.
It's silly to keep changing names.
To prove that AC is parallel to BD and AF is parallel to BE, we need to use the properties of intersecting chords in a circle.
Let's start by drawing the given diagram and labeling the points and angles as mentioned:
1. Draw two circles that intersect at points A and B, and label them as C₁ and C₂.
2. Draw a chord CD in one circle (C₁) that extends to meet the other circle (C₂) at points E and F, forming a straight line CDEF.
3. Extend the common chord AB to meet the line CF at point M.
4. Label the point of intersection of the lines AC and BD as P.
5. Label the point of intersection of the lines AF and BE as Q.
6. Given that M is the midpoint of CF, we can label CM as MF (as M is the midpoint).
7. Given angle CAF is 90 degrees, we can label that angle as 90 degrees.
Now, let's prove that AC is parallel to BD:
8. Since angle CAF is 90 degrees, and angle CAF and angle CAD (angle between chord CD and tangent at point A) are inscribed angles that subtend the same arc, angle CAD is also 90 degrees.
9. This means that CD is a diameter of circle C₁ (since it subtends a 90-degree angle at point A).
10. By the properties of diameters and chords, any chord that is perpendicular to a diameter is bisected by that diameter. Therefore, AM is congruent to MP.
11. Similarly, since CB is perpendicular to AB (as it subtends a 90-degree angle at point B), BP is also congruent to PA.
12. Therefore, we have congruent triangles APM and BPQ (by SSS congruence, as we have two pairs of congruent sides and a congruent included angle).
13. By congruent triangles, angle PAM is congruent to angle QBP (corresponding angles).
14. Since we know that angle CAD is 90 degrees, angle PAM and angle QBP sum up to 90 degrees.
15. By the converse of alternate interior angles, if two lines are cut by a transversal such that alternate interior angles are congruent, then the lines are parallel. Therefore, AC is parallel to BD.
Now, let's prove that AF is parallel to BE:
16. Since AC and BD are parallel lines (proved above), we can use the converse of alternate interior angles again. The alternate interior angles MAF and QBA (which sum up to 90 degrees) are congruent.
17. Similarly, since CB is perpendicular to AB, angle QEB (opposite angle of BAE) is also congruent to angle QBA.
18. Therefore, by the transitive property of congruence, angle MAF is also congruent to QEB.
19. By the converse of alternate interior angles, if two lines are cut by a transversal such that alternate interior angles are congruent, then the lines are parallel. Therefore, AF is parallel to BE.
Hence, we have proved that AC is parallel to BD and AF is parallel to BE.