# 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 tri deb and acb,
angle DBE=ANGLE ABC (COMMON)
ANGLE DEB =ANGLE ACB(EACH 90 )
THEREFORE  TRI DEB AND ACB ARE SIMILAR (BY AA RULE)
BE/BC=AC/DE
THEREFORE, BE/DE=AC/BC
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In the figure, AB ⊥ BC, DE ⊥ AC and GF ⊥ BC. Prove that ΔADE ~ ΔGCF
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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

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how is ✓ABC=✓DEB??
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In fig.DB L BC,DE L AB and AC L BC. Prove that BE/DE=AC/BC.

solve this fast
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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
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Hope it will help • 70
Ans • 0
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