In acute triangle ABC, we have \angle CAB = 34 ^\circ . E is the foot of the perpendicular from B to AC. A point H is on BE is chosen, and X is the foot of the perpendicular from H to AB. What is the measure (in degrees) of \angle EHX?
From X drop a perpendicular to intersect AC at Y.
XY ║ BX
Draw the diagram and it should be easy to see that EHX = 180-34 = 143
Sorry, that's
XY ║ BE
To find the measure of angle EHX, we can use the properties of triangles and angles. Let's break down the steps:
1. Draw the given acute triangle ABC, where angle CAB is 34 degrees.
2. Draw the perpendicular from B to AC, and label the foot as point E.
3. Draw a point H on BE.
4. Draw the perpendicular from H to AB, and label the foot as point X.
5. Now, we need to observe the relationships between the angles in triangle ABX and triangle EHX.
6. Triangle ABX has a right angle at X, so angle BAX is 90 degrees.
7. Triangle EHX also has a right angle at X, so angle EXH is 90 degrees.
8. Now, let's consider triangle ABH. Angle BAX is 90 degrees, and HAX is a straight line, so angle BAH must be 180 - 90 = 90 degrees.
9. Since angle EHB is a straight line, angle EHB + angle BAH = 180 degrees.
10. We know angle BAH is 90 degrees, so angle EHB must be 180 - 90 = 90 degrees as well.
11. Finally, in triangle EHX, angle EXH is 90 degrees, and angle EHB is 90 degrees. Therefore, angle EHX = angle EXH + angle EHB = 90 + 90 = 180 degrees.
So, the measure of angle EHX is 180 degrees.