Chapter 3.4, Problem 40E

Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

Chapter
Section

Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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In right triangle A B C , m ∠ C = 90 ° . Also, B C = a , C A = b , and A B = c . Construct the bisector of ∠ B so that it intersects C A ¯ at point D . Now construct D E ¯ perpendicular to A B ¯ with E on A B ¯ . In terms of a , b ,   and   c , find the length of E A ¯ .

To determine

To construct:

The bisector of B so that it intersects CA¯ at the point D and to construct DE¯ perpendicular to AB¯ with E on AB¯ in a right triangle ABC and to find the length of EA¯.

Explanation

Given:

In a triangle ABC,

mâˆ C=90Â°. Also, BC=a, CA=b, and AB=c

Property used:

(1) By Angle Bisector Theorem, angle bisector of any triangle divides the opposite sides of the angle to the ratio of the other two sides.

(2) If two angles and a non included side of one triangle are congruent to two angles and a non included side of a second triangle, then the triangles are congruent (AAS).

(3) Pythagorean Theorem:

The square of the length(c) of the hypotenuse of a right triangle equals the sum of squares of the lengths (a and b) of the legs of the right triangle; that is, c2=a2+b2.

Approach:

Consider the right triangle ABC with mâˆ C=90Â°, BC=a, CA=b, and AB=c.

Figure (1)

Construct the bisector of âˆ B so that it intersects CAÂ¯ at the point D.

Cut an arc from the point B that intersects the line BCÂ¯â€‰andâ€‰ABÂ¯ at points Pâ€‰andâ€‰Q. Now cut arcs with the same radius from points Pâ€‰andâ€‰Q. These two arcs intersect each other at a point R. Now draw a line by joining the points Bâ€‰andâ€‰R. This line intersects the line CAÂ¯ at a point D.

Figure (2)

Construct DEÂ¯ perpendicular to ABÂ¯ with E on ABÂ¯. Cut an arc from the point D that intersects line ABÂ¯ at points Sâ€‰andâ€‰T. Now cut arcs with the same radius from points Sâ€‰andâ€‰T. These two arcs intersect at a point U. Now draw the line by joining points Dâ€‰andâ€‰U that intersects ABÂ¯ on the point E. The line DEÂ¯ is perpendicular to ABÂ¯.

Figure (3)

Figure (4)

Now use Angle Bisector Theorem in Î”ABC.

ABÂ¯BCÂ¯=DAÂ¯CDÂ¯

Substitute CAÂ¯âˆ’CDÂ¯ for DAÂ¯ in the above equation.

ABÂ¯BCÂ¯=CAÂ¯âˆ’CDÂ¯CDÂ¯ABÂ¯BCÂ¯=CAÂ¯CDÂ¯âˆ’1

Now substitute a for BCÂ¯, b for CAÂ¯, and c for ABÂ¯ in the above equation.

caÂ¯=bCDÂ¯âˆ’1caÂ¯+1=bCDÂ¯c+aa=bCDÂ¯CDÂ¯=abc+a

Now substitute abc+a for CDÂ¯ and b for CAÂ¯ in DAÂ¯=CAÂ¯âˆ’CDÂ¯

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