Question

question number 28 please solve

Answer 1

TRIANGLES ADC AND ACB ARE SIMILAR SINCE

ANGLE CAD = ANGLE CAB........SAME ANGLE AREM EXTENDED

ANGLE ADC =90 = ANGLE ACB..GIVEN

HENCE

AD/AC = AC/AB

AC^2=AD*AB..............1

SIMILARLY TRIANGLES BCD AND BAC ARE SIMILAR SINCE

ANGLE CBD=ANGLE CBA

ANGLE BDC = 90 = ANGLE BCA ...HENCE

BC/BA = BD/BC

BC^2=BA*BD.................2

EQN.1/EQN.2...GIVES

AC^2/BC^2 = AD*AB/AB*BD=AD/BD...PROVED