In triangle ABC, points D,E,F are on sides BC,CA,AB respectively such that AD,BE,CF are angle bisectors of triangle ABC. The linesAD,BE,CF are concurrent at I,the incenter of triangle ABC. If Angle BAC=90, what is the measure (in degrees) of Angle EIF?
IN ABC, m<BAC =3x+9, m<ABC =8x+11, and m<BCA= 5x-8. Find the measures of <1,<2,<3. SHow your woek for solving x and finding each angle measure clearly.
IN ABC, m<BAC =3x+9, m<ABC =8x+11, and m<BCA= 5x-8. Find the measures of <1,<2,<3. SHow your woek for solving x and finding each angle measure clearly.
To find the measure of angle EIF, we need to use the properties of angle bisectors in a triangle.
Since angle BAC is a right angle, triangle ABC is a right triangle. Let's label the vertices as follows: angle B = 90 degrees, angle A = angle BAC, angle C = angle BCA.
We know that AD is the angle bisector of angle BAC. Therefore, angle DAB = angle DAC = angle A/2.
Similarly, BE is the angle bisector of angle ABC. Therefore, angle EBA = angle EBC = angle B/2 = 45 degrees.
CF is the angle bisector of angle BCA. Therefore, angle FCB = angle FCA = angle C/2 = 45 degrees.
Now, let's find the measure of angle EIF.
Since points D, E, F are on sides BC, CA, AB respectively, we have:
angle EIF = angle DEF + angle EDF.
angle DEF is equal to angle DAB + angle EBA = (angle A/2) + 45 degrees.
angle EDF is equal to angle DAC + angle FCB = (angle A/2) + 45 degrees.
Therefore, angle EIF = (angle A/2) + 45 degrees + (angle A/2) + 45 degrees
= angle A + 90 degrees.
Given that angle BAC is a right angle (90 degrees), we have:
angle EIF = 90 degrees + 90 degrees = 180 degrees.
Hence, the measure of angle EIF is 180 degrees.