$CD$ bisects $\angle AOB$, $DA$ and $DB$ are perpendiculars from point $D$ on the rays $OA$ and $OB$. show that $DA=DB$.
"

Given: $OD$ bisects $\angle AOB$ and $DA$ & $DB$ are perpendicular from point $D$ on the rays $OA$ and $OB$.

To do: To show that $DA=DB$.

We have

$\angle AOD = \angle BOD$ …… $( i)$ ….. as OD bisects $\angle AOB$ $( given)$

$\angle OAD = \angle OBD = 90^{o}$ …… (ii) ……. [DA and DB are perpendicular from point D on rays OA & OB $( given)$]

Now, let’s consider $\vartriangle ADO$ & $\vartriangle BDO$, we have

$\angle AOD = \angle BOD$ ….. from $( i)$

$OD = OD$ …… [common side to both the triangles]

$\angle OAD = \angle OBD$ …… from $( ii)$

$\therefore$ By ASA congruence, $\vartriangle ADO \cong \vartriangle BDO$

We know that if two or more triangles are congruent to each other, then all of their corresponding angles and sides are congruent as well. This property is known as Corresponding Parts of Congruent Triangles which is abbreviated as “C.P.C.T”. 

$\therefore DA = DB$