Question

Suggest a plan for proving that $\angle \mathbf{D}\cong \angle \mathbf{F}$.

Short answer

Plan for proving$\angle \mathbf{D}\cong \angle \mathbf{F}$is

Prove$\mathrm{\Delta }\mathbf{PEC}\cong \mathrm{\Delta }\mathbf{QEG}$ then$\overline{\mathrm{CP}}\cong \overline{\mathrm{GQ}}$ then prove$\mathrm{\Delta }\mathbf{DPC}\cong \mathrm{\Delta }\mathbf{FQG}$ and $\angle \mathbf{D}\cong \angle \mathbf{F}$.

Step-by-step solution

Step 1. Description of step.

Consider $\mathrm{\Delta }PEC$and $\mathrm{\Delta }QEG$in which$\overline{EC}\cong \overline{EG},\angle EPC\cong \angle EQG$and$\angle PEC\cong \angle QEG$then by AAS postulate width="111">ΔPEC≅ΔQEG.

Step 2. Description of step.

As$\mathrm{\Delta }PEC\cong \mathrm{\Delta }QEG$then by corresponding parts of congruent triangles, width="67">CP¯≅GQ¯.

Step 3. Description of step.

Consider$\mathrm{\Delta }DPC$ and$\mathrm{\Delta }FQG$ in which$\overline{DC}\cong \overline{FG},\angle DPC\cong \angle FQG$ and$\overline{CP}\cong \overline{GQ}$ then by SAS postulate $\mathrm{\Delta }DPC\cong \mathrm{\Delta }FQG$.

Step 4. Description of step.

As$\mathrm{\Delta }DPC\cong \mathrm{\Delta }FQG$ then by corresponding parts of congruent triangles, $\angle D\cong \angle F$.