Chapter 4.6, Problem 4CE

To determine

To prove: $\angle D\cong \angle F$

Explanation of Solution

Proof:

  

First prove that $\Delta PEC\cong \Delta QEG$

  $\overline{CE}=\overline{GE}\text{(Given)}$

As $\angle P=\angle Q=90°$ and two right angles are always congruent,

  $\angle P\cong \angle Q$

  $\angle PEC=\angle QEG\text{(Vertically}\text{opposite}\text{angles})$

So,

  $\Delta PEC\cong \Delta QEG$ (AAS congruence criteria)

(According to AAS congruence criteria, if two angles and the sides not included between them are equal to the two angles and the sides not included between them of the other triangle then the triangles are said to be congruent)

  $\overline{CP}\cong \overline{GQ}(i)$ (Corresponding parts of congruent triangles are congruent)

Now to prove, $\Delta DPC\cong \Delta FQG$

Consider $\Delta DPC,\Delta FQG$

As $\angle DPC=\angle FQG=90°$ and two right angles are always congruent,

  $\angle DPC=\angle FQG$

  $\overline{CP}=\overline{GQ}$ (Proved in $(i)$ )

  $\overline{DC}=\overline{GF}\text{(Given)}$

So,

  $\Delta DPC\cong \Delta FQG$ (RHS congruent criteria)

(According to RHS congruence criteria, if in a right-angled triangle, the hypotenuse and the side of one triangle are congruent to the hypotenuse and the side of other right-angled triangle then the triangles are said to be congruent)

Hence,

  $\angle D\cong \angle F$ (Corresponding parts of congruent triangles are congruent)