Question

IN THE GIVEN FIGURE, DB IS PERPENDICULAR TO BC, DE IS PERPENDICULAR TO AB AND AC IS PERPENDICULAR TO BC. PROVE THAT BE/DE= AC/BC.

Answer 1

Here is the answer to your question.

 



 

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

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Answer 2

thank you for asking the question.

although you don't give the picture but i know that question so i may help you out .

In the triangle DEB and triangle ACD.

angle DEB=Angle ABC (you can see if you have the picture).

anlge DEB = angle ACD (each is of 90)

therefor we can easily say that triangle DEB= triangle ACB  ( according to AA rule)

Hence BE/BC =AC/DE

 And then now we can have BE/DE = AC/BE