Similar Triangles

From the figure below:

5 thoughts on “Similar Triangles”

joshua_93 says:

April 23, 2012 at 4:39 pm

Let ∠ACB be a;
Since ∠EAF = 90, then ∠CAB= 90;
∠CED = 180-90-a
= 90-a (angle sum of triangle = 180)
∠CBA = 180-90-a
= 90-a (angle sum of triangle = 180)
∠FEA = 90-a (vertically opposite)
∠EFA = 180-(90-a)
= 90+a (angle sum of triangle=180)
∠AED = 180-(90-a)
= 90+a (supplementary)

Proof for similarity between Δ DEA ||| Δ ADF;
A = ∠EFA = ∠DEA
A = ∠EDA = ∠EDA

Thus, Δ DEA ||| Δ ADF

Reply
1. hedy says:
  
  April 24, 2012 at 1:39 pm
  
  Josh, please check on the angle EFA 🙂
  
  Reply
  1. Jonathan says:
    
    April 24, 2012 at 10:55 pm
    
    Angle EFA is 180 – ( 90 – a ) – 90 or a.
    
    Reply
    1. joshua_93 says:
      
      April 25, 2012 at 12:17 pm
      
      lol. yup that’s what i mean. hahah. i miscalculate it.
      Point D seems to be the midpoint of line CB. Thus:
      ∠EAD = a,
      therefore ∠EFA = ∠EAD
      
      Then, we can get:
      ∠FAD = 90+a;
      ∠AED = 180-(90-a)
      = 90+a (supplementary);
      ∠FAD =∠AED
      
      Proof for similarity between Δ DEA ||| Δ ADF;
      A = ∠FAD =∠AED
      A = ∠EDA = ∠EDA
      
      Thus, Δ DEA ||| Δ ADF 🙂
      
      uggh bener gaakk sih?
      
      Reply
      1. hedy says:
        
        May 1, 2012 at 9:24 am
        
        Thanks ya both Joshua and Jonathan. Both are correct 🙂
        Check on my latest question / quiz after this 🙂