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.

 In the figure, AB ⊥ BC, DE ⊥ AC and GF ⊥ BC. Prove that ΔADE ~ ΔGCF

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
 

how is ✓ABC=✓DEB??

In fig.DB L BC,DE L AB and AC L BC. Prove that BE/DE=AC/BC.

AC and BC are perpendicular to BC, so AC and BC are parallel to each other. Now, in triangle ACB and triangle BAD angle CAB = angle ABD (Alternate interior angles) Also, angle ACB = angle DEB (90 degree each) Therefore BE/DE = AC/BC Hence Proved

Hope it will help

Ans

aaaaaaaaaaa

answers

Hope u like it

Is it helpful