Question

In figure, $ ∆ \mathrm{ARO} \cong ∆ \mathrm{\_\_\_\_\_\_}$

Answer 1

Finding which triangle is congruent to triangle $\mathrm{ARO}$:

Given: $\mathrm{AO}=\mathrm{OP}=2.5\mathrm{cm},\angle \mathrm{ARO}=\angle \mathrm{PQO}=55°$

$$\begin{gathered} \angle \mathrm{AOR}=\angle \mathrm{QOP}(\mathrm{vertically}\mathrm{opposite}\mathrm{angles}) \\ \mathrm{AO}=\mathrm{OP}\left(\mathrm{Given}\right) \\ \angle \mathrm{ARO}=\angle \mathrm{PQO}\left(\mathrm{Given}\right) \end{gathered}$$

$\therefore ∆\mathrm{ARO}\cong ∆\mathrm{PQO}(\mathrm{by}\mathrm{ASA}\mathrm{congruency})$

Hence, the required answer is $∆\mathrm{ARO}\cong ∆\mathrm{PQO}.$