i already proved the triangles.i need help with finding ratio of BXC to rectangle ABCD. it is given that CM=3MB
Produce a straight line where Y lies on AB, Z lies on DC, and YXZ//AD//BC.
CM = 3DM, hence
CM = 0.75*DC = 0.75*AB
AB / CM
= AB / 0.75AB
= AX / CX
= BX / MX
triangle AYX similiar to CZX (AAA), therefore
AY / CZ = AX / CZ = 4/3
Therefore CZ = BY = (3/7)*AB
area of triangle BXC = BC*BY/2
area of rectangle ABCD = (BC)(AB)
Hence, the ratio concerned
regarding to BY = (3/7)*AB :
first have a look at AY:BY again. it is 4:3. it means that WHEN AY is 4, BY is 3.....when AY is 20, BY is 15, etc.
it tells us that WHEN the whole length of AB is EQUALLY divided into 7 PARTS, then AY constitutes 4 parts, and BY consitutes 3.
therefore AY:BY = 4:3, AY:AB = 4:7, BY:AB = 3:7
These numbers do NOT have units (cm, mm, etc). it is all about ratios.
if you want to prove with real lengths, consider AY = 20cm, then BY = 15cm, and hence AB = 35cm. you will get the same ratios as aforesaid.
to shorten the calculation, it is not really necessary to have point Z and line XZ but you need Y and line BY which is equivalent to the height of triangle required to calculate its area. simply use the ratio of AX:CX to work out the ratios of AY:BY and AB:BY, and eventually present BY in terms of AB.