A Andy005 Member Joined Mar 6, 2017 Messages 57 Location Gosford Gender Male HSC 2018 Aug 22, 2017 #1 https://drive.google.com/open?id=0B_RlXPQTW1cRQUlmYURxeWZjZkE Any help would be greatly appreciated. Thanks Last edited: Aug 23, 2017
https://drive.google.com/open?id=0B_RlXPQTW1cRQUlmYURxeWZjZkE Any help would be greatly appreciated. Thanks
K K-dogg New Member Joined Apr 26, 2016 Messages 1 Gender Undisclosed HSC N/A Aug 24, 2017 #2 From the diagram setup it seems likely you'll be using the alternate segment theorem. I'll start from BDA = MAB (AST)
From the diagram setup it seems likely you'll be using the alternate segment theorem. I'll start from BDA = MAB (AST)
pikachu975 Premium Member Joined May 31, 2015 Messages 2,739 Location NSW Gender Male HSC 2017 Aug 24, 2017 #3 let MAB = a, BAD = b BDA = MAB = a (angles in alternate segments) BCA = BDA = a (angles in the same segment) NAC = BCA = a (alternate angles, MN || BC) CDA = NAC = a (angles in alternate segments) BDA = CDA = a therefore AD bisects BDC
let MAB = a, BAD = b BDA = MAB = a (angles in alternate segments) BCA = BDA = a (angles in the same segment) NAC = BCA = a (alternate angles, MN || BC) CDA = NAC = a (angles in alternate segments) BDA = CDA = a therefore AD bisects BDC