In the given figure, OA = OB and OP = OQ. Prove that (i) PX = QX, (ii) AX = BX.

Solution:

Proof:

In $\triangle \mathrm{OQA}$ and $\triangle \mathrm{OPB}$, we have :

$\mathrm{OQ}=\mathrm{OP} \quad$ (Given)

$\mathrm{OA}=\mathrm{OB} \quad$ (Given)

$\angle \mathrm{AOQ}=\angle \mathrm{BOP} \quad$ (Common)

$\triangle \mathrm{OQA} \cong \mathrm{OPB} \quad$ (SAS criterion)

$\angle O A Q=\angle O B P \quad$ (Corresponding angles of congruent triangles)

Now, consider triangles BQX and APX.

Given :

$\mathrm{OA}=\mathrm{OB}$

$\mathrm{OP}=\mathrm{OQ}$

$\therefore \mathrm{OA}-\mathrm{OP}=\mathrm{OB}-\mathrm{OQ}$

$\Rightarrow \mathrm{AP}=\mathrm{BQ}$

Further, $\angle \mathrm{BXQ}=\angle \mathrm{AXP} \quad$ (Vertically opposite angles)

Also, we have proven that $\angle Q B X=\angle P A X$.

$\Delta \mathrm{BQX} \cong \Delta \mathrm{APX} \quad$ (AAS criterion)

$\therefore \mathrm{PX}=\mathrm{QX} \quad$ (corresponding sides of congruent triangles)

Also, $\mathrm{AX}=\mathrm{BX} \quad$ (corresponding sides of congruent triangles)

Hence, proved.