Q24. For Exercises 23-27 write proofs... [FREE SOLUTION]

91影视

Geometry

Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

$Math Studyset 91影视 Explanations$ Math

Student Edition 路 227 pages

ISBN: 9780395977279

Chapter 4: Q24. (page 157)

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: $\overset{鈫�}{D P}$ bisects $鈭� A D E; \overset{鈫�}{E P}$ bisects $鈭� D E C$
Prove:width="26" height="24" role="math">BP鈫� bisects $鈭� A B C$

Short Answer

Expert verified

First use theorem 4-7, 鈥溾€淚f the point lies on the bisector of an angle, then the point is equidistance from the sides of the angle鈥�. Since, point $P$ lies on the angle bisector of $鈭� A D E$ , so point $P$ is equidistance from sides $\overset{炉}{A D} and \overset{炉}{D E}$ . Similarly point $P$ lies on the angle bisector of $鈭� D E C$ , so point $P$ is equidistance from sides $\overset{炉}{D E} and \overset{炉}{E C}$ . Point $P$ is equidistant from sides $\overset{炉}{A D} and \overset{炉}{D E}$ and also $\overset{炉}{D E} and \overset{炉}{E C}$

Therefore, point $P$ is equidistant from sides $\overset{炉}{A D} and \overset{炉}{C E}$ , or one can say that point $P$ is equidistant from sides $\overset{炉}{A B} and \overset{炉}{B C}$

Use Theorem 4-8, 鈥淚f a point is equidistance from the sides of an angle, then the point lies on the bisector of the angle鈥�

Since point $P$ is equidistant from sides $\overset{炉}{A B} and \overset{炉}{B C}$ then, point $P$ lies on the bisector of $鈭� A B C$ or $\overset{鈫�}{B P}$ bisects $鈭� A B C$

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 $鈭� A D E$ , so using theorem 4-7 鈥淚f the point lies on the bisector of an angle, then the point is equidistance from the sides of the angle鈥� point $P$ is equidistance from sides $\overset{炉}{A D} and \overset{炉}{D E}$ .

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

Since, point $P$ lies on the angle bisector of $鈭� D E C$ , so using theorem 4-7 鈥淚f the point lies on the bisector of an angle, then the point is equidistance from the sides of the angle鈥� point $P$ is equidistance from sides $\overset{炉}{D E} and \overset{炉}{E C}$ .

Step 3. Show that BP→ bisects ∠ABC

Point $P$ is equidistant from sides $\overset{炉}{A D} and \overset{炉}{D E}$ and also $\overset{炉}{D E} and \overset{炉}{E C}$

Therefore, point $P$ is equidistant from sides $\overset{炉}{A D} and \overset{炉}{C E}$ , or one can say that point $P$ is equidistant from sides $\overset{炉}{A B} and \overset{炉}{B C}$

Use Theorem 4-8, 鈥淚f a point is equidistance from the sides of an angle, then the point lies on the bisector of the angle鈥�

Since point $P$ is equidistant from sides $\overset{炉}{A B} and \overset{炉}{B C}$ then, point $P$ lies on the bisector of $鈭� A B C$ or $\overset{鈫�}{B P}$ bisects $鈭� A B C$