How will I use this proportion?
Follow Math Help Forum on Facebook and Google+
Originally Posted by Idea CD meets BE at F AC = BF = a BE = b We let h and k be the distances from C and D to the side BE 4 = area BCF - area BDF = 1/2 (h - k) a Use the proportion $\displaystyle \frac{k}{b-a}=\frac{h}{b}$ How did you find this proportion?
EDF and EAB are similar triangles therefore $\displaystyle \frac{\text{EF}}{\text{EB}}=\frac{k}{h}$
I beg your pardon. I am solving nearly 70 questions a day. My brain stopped. ABE=9 CDA=4 We don't know the area of CFE. Area of BECD is 5. To use the above ratio we need the area of FCE or BDE or DFE
area BCD = 4 = area ADC area BCE = 9 = area ABE
Last edited by Idea; May 30th 2017 at 03:59 AM.
ADC and DFE are similar. CD/AB=2/3 DC/DF=2 ABC=ACE DFE=2 CDE=2 ABC=6 Is that right?
Originally Posted by kastamonu adc and dfe are similar. Cd/ab=2/3 dc/df=2 abc=ace dfe=2 cde=2 abc=6 is that right? dfe = 1 cde = 2 bdf = 2 abc= 6
I made a typo DFE is 1. Many Thanks.