Question

pls solve 3c i will mark as brainliest

Answer 1

Hii

Given: In ∆ABC and ∆DBE, AC⊥BC, BD ⊥BC and DE ⊥AB.

 To prove BE/DE = AC/BC

Proof: In ∆ABC and ∆DBE,

‹c = ‹deb 90°

angle abc + angle DBE = 90°

Angle BDE + angle DBE = angle BD + Angle DBE

equal to angle abc equal to angle BDE

Triangle ABC congrance Triangle DBE

equal to AB by BD equal to AC by BE equal to DC by DE are common side

equal to AC by BC equal to BE by DE

Now

BC×BE= DC×AC

Hence proved

Hope thse would helps u ✌

Answer 2

Given :

In Δ DEB and Δ ABC ,

∠DEB = ∠ACB [ 90° each ]

∠DBC = 90°

= > ∠DBE + ∠ABC = 90

= > ∠ DBE = 90 - ∠ABC

= > ∠ DBE = ∠ CAB

Δ DEB ≈ Δ ABC [ A.A ]

Hence :

BE / DE = AC / BC

= > BE × BC = DE × AC

Hence Proved .