Question

solve it fast as fast as you can​

Answer 1

AB = a  

Given that CD = 2AB = 2a.  

If a line BI is drawn from B to CD, parallel to AD, it will make right angle at BIC.  

Since AB = DI, IC = 2a – a = a.  

In ΔBIC and ΔBHE,  

∠HBE = ∠IBC

∠BHE = ∠BIC

∠ICB = ∠HEB  (HE is parallel to IC and EC is the tangent cutting both parallel lines.)

ΔBIC ~ ΔBHE

BE/BC = HE/IC

3x/7x = HE/a

HE = 3a/7

EF = FH + HE = a + 3a/7 = 10a/7

7EF = 10a = 10AB

Hence proved.