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: $\overrightarrow{\mathit{D}\mathit{P}}$bisects $\mathbf{\angle }\mathit{A}\mathit{D}\mathit{E}\mathbf{;}\overrightarrow{\mathit{E}\mathit{P}}$bisects $\mathbf{\angle }\mathit{D}\mathit{E}\mathit{C}$

Prove:width="26" height="24" role="math">BP→ bisects$\mathbf{\angle }\mathit{A}\mathit{B}\mathit{C}$

Short answer

First use theorem 4-7, ““If the point lies on the bisector of an angle, then the point is equidistance from the sides of the angle”. Since, point $P$lies on the angle bisector of $\angle ADE$, so point $P$is equidistance from sides $\overline{AD}\text{and}\overline{DE}$. Similarly point $P$lies on the angle bisector of $\angle DEC$, so point $P$is equidistance from sides $\overline{DE}\text{and}\overline{EC}$. Point $P$is equidistant from sides $\overline{AD}\text{and}\overline{DE}$and also $\overline{DE}\text{and}\overline{EC}$

Therefore, point $P$is equidistant from sides $\overline{AD}\text{and}\overline{CE}$, or one can say that point $P$is equidistant from sides $\overline{AB}\text{and}\overline{BC}$

Use Theorem 4-8, “If a point is equidistance from the sides of an angle, then the point lies on the bisector of the angle”

Since point $P$is equidistant from sides $\overline{AB}\text{and}\overline{BC}$then, point $P$lies on the bisector of$\angle ABC$ or $\overrightarrow{BP}$bisects$\angle ABC$

Step-by-step solution

Step 1. Show that P is equidistance from sides AD¯ and DE¯

Since, point$P$ lies on the angle bisector of $\angle ADE$, so using theorem 4-7 “If the point lies on the bisector of an angle, then the point is equidistance from the sides of the angle” point$P$ is equidistance from sides $\overline{AD}\text{and}\overline{DE}$.

Step 2. Show that P is equidistance from sides DE¯ and EC¯

Since, point$P$ lies on the angle bisector of $\angle DEC$, so using theorem 4-7 “If the point lies on the bisector of an angle, then the point is equidistance from the sides of the angle” point$P$ is equidistance from sides $\overline{DE}\text{and}\overline{EC}$.

Step 3. Show that BP→ bisects ∠ABC

Point $P$is equidistant from sides $\overline{AD}\text{and}\overline{DE}$and also $\overline{DE}\text{and}\overline{EC}$

Therefore, point $P$is equidistant from sides $\overline{AD}\text{and}\overline{CE}$, or one can say that point $P$is equidistant from sides $\overline{AB}\text{and}\overline{BC}$

Use Theorem 4-8, “If a point is equidistance from the sides of an angle, then the point lies on the bisector of the angle”

Since point $P$is equidistant from sides $\overline{AB}\text{and}\overline{BC}$then, point $P$lies on the bisector of$\angle ABC$ or $\overrightarrow{BP}$bisects$\angle ABC$