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 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¯.

Given:

In a triangle ABC,

m∠C=90°. 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 m∠C=90°, BC=a, CA=b, and AB=c.

Figure (1)

Construct the bisector of ∠B so that it intersects CA¯ at the point D.

Cut an arc from the point B that intersects the line BC¯ and AB¯ at points P and Q. Now cut arcs with the same radius from points P and Q. These two arcs intersect each other at a point R. Now draw a line by joining the points B and R. This line intersects the line CA¯ at a point D.

Figure (2)

Construct DE¯ perpendicular to AB¯ with E on AB¯. Cut an arc from the point D that intersects line AB¯ at points S and T. Now cut arcs with the same radius from points S and T. These two arcs intersect at a point U. Now draw the line by joining points D and U that intersects AB¯ on the point E. The line DE¯ is perpendicular to AB¯.

Figure (3)

Figure (4)

Now use Angle Bisector Theorem in ΔABC.

AB¯BC¯=DA¯CD¯

Substitute CA¯−CD¯ for DA¯ in the above equation.

AB¯BC¯=CA¯−CD¯CD¯AB¯BC¯=CA¯CD¯−1

Now substitute a for BC¯, b for CA¯, and c for AB¯ in the above equation.

ca¯=bCD¯−1ca¯+1=bCD¯c+aa=bCD¯CD¯=abc+a

Now substitute abc+a for CD¯ and b for CA¯ in DA¯=CA¯−CD¯