Question

In fig. chord $AB$ bisects chord $CD$ at $O$, Prove that –
$O{A}^2=DO\cdot OC$

Answer 1

Given : chord $AB$ bisects chord $CD$ at $O$
$CO=OD$ …..(i)
Join $BC$
$\angle CAB$ and $\angle BDC$, arc angles of same segment so will be equal
$\angle CAB=\angle BDC$ …..(ii)
In $\Delta COA$ and $\Delta DOB$
$\angle CAO=\angle BDO$ (from equation (ii))
$CO=OD$ (given eqn. (i)
$\angle COA=\angle DOB$ (vertically opposite angles)
By ASA congruence rule.
$\Delta COA=\Delta DOB$
Thus, corresponding sides of congruent triangle are same,
$OA=OB$
$OB=OC$ [$\because OB$ and $OC$, arc radius of circle]
$OA=OC$
$(OA{)}^2=(OC{)}^2$
$(OA{)}^2=OC\times OC$
$O{A}^2=OC\times OD$ $[OC=OD]$
or $O{A}^2=DO\times OC$