March 21st, 2013, 08:18 PM  #1 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Find Angle A
[attachment=0:76gf9of6]Find Angle A.JPG[/attachment:76gf9of6]

March 22nd, 2013, 02:43 AM  #2 
Senior Member Joined: Feb 2010 Posts: 714 Thanks: 151  Re: Find Angle A
area of triangle ADE = (1/2)(AD)(AE)(sin A) and area of triangle ABC = (1/2)(AB)(AC)(sin A) Therefore (1/2)(AD)(AE)(sin A) = (1/4)(AB)(AC)(sin A) This simplifies to 2(AD/AC) = AB/AE which is 2 cos A = 1/cos A Thus cos A = 1/sqrt(2) and A = 45 degrees 
March 22nd, 2013, 05:29 AM  #3 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: Find Angle A
very good , you got it 
March 23rd, 2013, 12:47 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 21,034 Thanks: 2269 
Let EF be an altitude of triangle ADE. Triangles BEC, EFD are similar and in ratio BE/EF. Triangles BEA, EFA are similar and in ratio BE/EF = AE/AF. ?EFD + ?EFA = (?BEC + ?BEA)/2 (given), so AF² = AE²/2. By Pythagoras, AE² = AF² + EF², so AF = EF. Hence angle EAF = 45°. 
March 23rd, 2013, 08:41 PM  #5 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: Find Angle A
skipjack :you always have a very good and unusual solution 

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