Euclid II-17. (pg 64)
- Author:
- Emily Crum
Explanation:
create the Circle BCD with center E. Create pt A outside the circle.
Join AE and use this as the radius of circle FA.
Create CF such that it is perpendicular to AE.
Join EF and AB. AB is a straight line touching circle BCD.
EF=EA because they are both radii of the larger circle. angle FEA= angle FEA. BE=CE because they are both radii of the smaller circle. Thus by SAS, triangle FCE = triangle BAE. Thus CF=BA and angle ECF= angle ABE.
But, angle ECF is a right angle (as given by earlier conditions where CF is perpendicular to AE).
Thus angle ABE must also be a right angle.
With EB as radius, the straight line drawn perpendicular to the radius, from its extremity, touches the circle. Thus, AB touches circle BCD.