Question

For Exercises 23-27 write proofs in paragraph form. (Hint: You can use theorems from this section to write fairly short proofs for Exercises 23 and 24.)

Given: Plane $\mathit{M}$is the perpendicular bisecting plane of $\overline{\mathit{A}\mathit{B}}$, (That is, $\overline{\mathit{A}\mathit{B}}\mathbf{\perp }$plane $\mathit{M}$and $\mathit{O}$is the midpoint of$\overline{\mathit{A}\mathit{B}}$)

Prove: $$\begin{gathered} \mathrm{a}.\overline{\mathit{A}\mathit{D}}\cong \overline{\mathit{B}\mathit{D}} \\ \mathrm{b}.\overline{\mathit{A}\mathit{C}}\cong \overline{\mathit{B}\mathit{C}} \\ \mathrm{c}.\mathbf{\angle }\mathit{C}\mathit{A}\mathit{D}\mathbf{\cong }\mathbf{\angle }\mathit{C}\mathit{B}\mathit{D} \end{gathered}$$

Short answer

1. Point $D$lies on the plane $M$which is perpendicular bisector for line segment $\overline{AB}$and $O$is its midpoint. So $\overline{OD}$is perpendicular bisector of $\overline{AB}$which implies point $D$is equidistance from endpoints $A$and $B$. Thus$\overline{AD}\cong \overline{BD}$
2. Point$C$lies on the plane $M$which is perpendicular bisector for line segment $\overline{AB}$and $O$is its midpoint. So $\overline{OC}$is perpendicular bisector of $\overline{AB}$which implies point $C$is equidistance from endpoints $A$and $B$. Thus$\overline{AC}\cong \overline{BC}$
3. In width="48" height="19" role="math">ΔCADand $\mathrm{\Delta }CBD$, using above two parts $\overline{AD}\cong \overline{BD}$and $\overline{AC}\cong \overline{BC}$. Also, $\overline{CD}$is common in both triangles, so by reflexive property it is congruent to itself. Thus, by SSS (Side-Side-Side) Postulate $\mathrm{\Delta }CAD\cong \mathrm{\Delta }CBD$Since, corresponding parts of congruent triangles are congruent, so$\angle CAD\cong \angle CBD$

Step-by-step solution

Part a. Step 1. Observation from image.

Point $D$lies on the plane $M$which is perpendicular bisector for line segment $\overline{AB}$and $O$is its midpoint. So $\overline{OD}$ is perpendicular bisector of$\overline{AB}$

Part a.  Step 2. Show that AD¯≅BD¯.

Since, $\overline{OD}$is perpendicular bisector of $\overline{AB}$, so point $D$is equidistance from endpoints $A$ and $B$. Thus$\overline{AD}\cong \overline{BD}$

Part b. Step 1. Observation from image.

Point $C$lies on the plane $M$which is perpendicular bisector for line segment $\overline{AB}$and $O$is its midpoint. So $\overline{OC}$ is perpendicular bisector of$\overline{AB}$

Part b. Step 2. Show that AD¯≅BD¯.

Since,$\overline{OC}$ is perpendicular bisector of $\overline{AB}$, so point$C$ is equidistance from endpoints$A$ and $B$. Thus$\overline{AC}\cong \overline{BC}$

Part c. Step 1. Show that ΔCAD≅ΔCBD.

In$\mathrm{\Delta }CAD$ and $\mathrm{\Delta }CBD$, using above two parts$\overline{AD}\cong \overline{BD}$ and$\overline{AC}\cong \overline{BC}$

Also,$\overline{CD}$is common in both triangles, so by reflexive property it is congruent to itself.

Thus, by SSS (Side-Side-Side) Postulate$\mathrm{\Delta }CAD\cong \mathrm{\Delta }CBD$

Part c. Step 2. Show that ∠CAD≅∠CBD.

Since, corresponding parts of congruent triangles are congruent and$\mathrm{\Delta }CAD\cong \mathrm{\Delta }CBD$

So,$\angle CAD\cong \angle CBD$