Plane Geo (II)

• bonkysleuth

  13 Dec 08, 00:32

  In the diagram, ABCD is Y is te midpoint of AB. XYD is a straight line and Z is the point of intersection of DB and YC. Given that XB is paralle to YC, prove that

   

  (a) triangle YDC is similar to triangle XYB

  (b) DY = 2 YX

  (c) DZ = 2 ZB

   

  I will provide you with my full solution before the sketch itself(which I drew using the paint programme). So pardon me for the bad drawing!

   

  For (a)

   

  since XB // YC

   

  angleDYC = angle YXB (corres.)

  angle YCD = angle XBY(corres. angle)

  Hence the triangles are similar (angle-angle-angle similarity)

   

  For (b)

  since YC //XB, and Y is midpoint of AB

   

  YC : XB = 2:1 (midpoint theorem)

  since triangle YDC and triangle XYB are similar,

   

  YC:XB = DY:YX = 2:1

   

  .: DY = 2 YX (proven)

   

  For (c)

  triangle YDC is similar to triangle XYB(from(b))

  triangle XDB is similar to triangle YDZ

   

  from (b), we know that

  YD:XY = 2:1

   

  YD:XD = YD : (YD+XY) = 2:3

  XD:YD = DB:DZ = (DZ+ZB):DZ

                         = 3:2

   

  .:ZB = 1

  Hence DZ = 2 ZB

  

  Hang on, there's another one which I want to check whether the method is valid. (: thanks for your patience.

• bonkysleuth

  13 Dec 08, 00:43

  In the trapezium ABCD, AD is parallel to BC, BC = 3AD and the diagonals AC and BD meet at E. The line through A is drawn parallel to DB to meet the extended line of CB at F. Prove that

   

  (a) FB = AD

  (b) EC = 3AE

   

  Again, the solutions first.

   

  (a) Triangle AED is similar to triangle CEB

   

  AD : CB = 1:3

  CB:BF = CE:EA = CE: DA(intercept theorem)

   

  since BF = DA based on above equation, FB = AD

   

  (b) Since triangle AED is similar to triangle CEB

   

  CB:BF = CE:EA=CB:AD = 3:1

   

  hence EC = 3 AE(proven)

   

  that's all. thank you once again.