Plane Geo (II)

bonkysleuth SG

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

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 SG

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.