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
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