Question

solve this question Pls ​

Answer 1

Answer:

in ∆ABG & ∆DCB we have ,

angle AGB = angle DBC (90° each)

angle GBA = angle BCD ( As GB is parallel to DC , so, alternate angles )

so,

∆ABG ~ ∆DCB (AA right ∆ propety)

now for part second ,

similary we can prove that ,

angle DBC = Angle ABE

angle EAB = angle angle CDB ( bcz ∆ABG ~ ∆DCB and angle EAB = angle GAB)

so,

∆BCD ~ ∆ BEA

so,

BC/BD = BE/AB (Proved)