Question

$ \mathrm{In} ∆\mathrm{ABC}, \mathrm{AB}=\mathrm{AC} \mathrm{and} \mathrm{AP}=\mathrm{AQ}.$ Prove that $ \mathrm{CP}=\mathrm{BQ}$.

Answer 1

Prove the required condition:

Given: $\mathrm{AB}=\mathrm{AC}\mathrm{and}\mathrm{AP}=\mathrm{AQ}$

$$\begin{gathered} \mathrm{In}∆\mathrm{ABC}, \\ \mathrm{AB}-\mathrm{AP}=\mathrm{AC}-\mathrm{AQ} \\ \mathrm{BP}=\mathrm{CQ} \end{gathered}$$

Now in $∆\mathrm{BCP}$ and $∆\mathrm{BCQ}$

$\mathrm{BP}=\mathrm{CQ}$

$\angle \mathrm{BCP}=\angle \mathrm{BCQ}$ (Common angles)

$\mathrm{BC}=\mathrm{BC}$ (Common side)

Hence $∆\mathrm{BCP}\cong \mathrm{BCQ}$ (by SAS congruency)

Hence $\mathrm{CP}=\mathrm{BQ}$ (corresponding sides of congruence triangles)

Hence, proved that $\mathrm{CP}=\mathrm{BQ}$.