1. In the diagram below, if AD=30, what is BC?
2. We know BE=CE and AE=2DE. What is AD/BC?
3. Given that AF=\(4sqrt(3)\) and FC=5sqrt(3) ,what is BC?
+128406 First one
Let the intersection of AD and BC = E
Triangle AEB is similar to Triangle DEC
Let DE = x and let AE = 30 - x
So
AE / AB = DE / DC
( 30 - x) / 20 = x / 4 cross-multiply
4 (30 - x) = 20x simplify
120 - 4x = 20x add 4x to both sides
120 = 24x divide both sides by 24
5 = x = DE
So....CE = sqrt [ DE^2 - DC^2] = sqrt [ 5^2 - 4^2 ] = sqrt [ 9] = 3
And BE is 5 times this = 15
So....BC = BE + EC = 15 + 3 = 18
+128406 Second one
Triangle ABE similar to triangle ECD
So
AE / BE = DE / CD
2DE / BE = DE / CD ......so......
2DE / DE = BE / CD
2 = BE / CD
So BE = 2CD ⇒ CD = BE / 2 = EC / 2
AD = sqrt ( AE^2 + DE^2] = sqrt [ (2DE)^2 + DE^2] = sqrt(5)DE
And
DE = sqrt [ EC^2 + (EC/2)^2] = EC sqrt(5) / 2 = BEsqrt(5)/2
So
DE = BE sqrt(5) / 2
2DE /sqrt(5) = BE
So
BC = 2BE
So
2BE = 4DE / sqrt(5) = BC
So
AD / BC = sqrt (5) DE / [ 4DE / sqrt (5) ]
AD /BC = sqrt (5) * sqrt(5) DE / [ 4DE ] = 5DE / [ 4 DE ] = 5 / 4
