Question

please solve this......​

Answer 1

given: DB⊥BC, DE⊥AB and AC⊥BC.

to prove: BE/DE = AC/BC

proof: in ΔDEB and ΔBCA

               ∠DEB = ∠BCA (90°)

               ∠DBE = ∠BAC (a.i.a)

              ∴ΔDEB≈ΔBCA (AA rule)

              ∴BE/AC = DE/BC ( corresponding sides of similar triangles are                    .                                             proportional)

               ∴ BE/DE = AC/BC

HENCE PROVEN!!!!!!!!!!!