Tri-Similarity (Posted on 2010-10-17)

	Submitted by Bractals
Rating: 3.0000 (1 votes)
Solution:	(Hide)
Let Q, R, and S be the intersections of ray AP and side CD, line AC and diagonal BD, and line AD and diagonal CE respectively. Using the similarity of triangles ABC, ACD, and ADE: /BCD = /BCA + /ACD = /CDA + /ADE = /CDE |CB| |CA| |DA| |DC| ------ = ------ = ------ = ------ |CD| |DA| |EA| |DE| Therefore,  ΔBCD ~ ΔCDE. /RDC = /BDC = /CED = /SED /DCR = /DCA = /EDA = /EDS Therefore,  ΔRCD ~ ΔSDE. |RC| |CD| |AC| |AR| + |RC| ------ = ------ = ------ = ------------- |DS| |DE| |DA| |DS| + |SA| or |AR||DS| = |RC||SA| (1) By Ceva's theorem, |AR||CQ||DS| = |RC||QD||SA| (2) Combining (1) and (2) gives, |CQ| = |QD| Thus, ray AP bisects side CD.

	Subject	Author	Date
	possible solution	broll	2010-10-19 06:47:11
	re(8): Question?	Bractals	2010-10-19 01:08:26
	re(7): Question?	broll	2010-10-18 19:32:37
	re(6): Question?	ed bottemiller	2010-10-18 18:17:34
	re(5): Question?	Bractals	2010-10-18 15:59:52
	re(4): Question?	broll	2010-10-18 15:02:39
	re(3): Question?	Bractals	2010-10-18 14:27:16
	re(2): Question?	broll	2010-10-18 03:53:25
	re: Question?	Bractals	2010-10-17 23:57:48
	Question?	broll	2010-10-17 16:15:11