Euclid II-17. (pg 64)
- 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.