Question

1. point h is the intersection of three medians. if bh=15, what is he and be?
2. given i in the incenter, find izn.

Explanation:

Problem 6:

Step1: Recall the property of the centroid of a triangle. The centroid (H) divides each median into a ratio of 2:1, where the distance from the vertex to the centroid is twice the distance from the centroid to the midpoint of the side. So, \( BH:HE = 2:1 \).

Given \( BH = 15 \), let \( HE = x \). Then, \( \frac{BH}{HE}=\frac{2}{1} \), so \( \frac{15}{x}=\frac{2}{1} \). Cross - multiplying gives \( 2x = 15 \), so \( x=\frac{15}{2}=7.5 \).

Step2: To find \( BE \), we know that \( BE=BH + HE \). We already found \( BH = 15 \) and \( HE = 7.5 \), so \( BE=15 + 7.5=22.5 \).

Step1: Recall the property of the incenter of a triangle. The incenter (I) is the intersection of the angle bisectors of the triangle. So, \( \angle ZIY\) is related to the angle bisector. We know that \( \angle ZIX = 74^{\circ}\) and \( \angle YIX=22^{\circ}\), and we want to find \( \angle IZN \). Wait, maybe there is a typo and it's \( \angle IZY \) or \( \angle IZN \) related to the right angle? Wait, looking at the triangle, if \( ZN \) is perpendicular to \( YX \) (assuming \( ZN \) is an altitude, but since I is the incenter, maybe we can use the fact that the incenter is equidistant from the sides and the angle bisector property. Wait, another approach: The sum of angles in a triangle and the angle bisector. Wait, maybe the triangle \( ZIY \) has some right angle? Wait, if \( ZN \) is perpendicular to \( YX \), then \( \angle ZNY = 90^{\circ}\). The incenter I, so \( \angle ZIY=180^{\circ}-(74^{\circ}+ 22^{\circ})=84^{\circ}\)? No, wait, the incenter angle formula: In a triangle, the measure of the angle formed by two angle bisectors is \( 90^{\circ}+\frac{1}{2}\) the measure of the original angle. Wait, maybe we need to find \( \angle IZN \). Since \( ZN \) is perpendicular to \( YX \) (assuming), and \( I \) is the incenter, \( \angle IZN = 90^{\circ}-\angle ZIX \)? Wait, no, let's re - examine. Wait, the angle at \( Z \): if \( \angle ZIX = 74^{\circ}\) and \( \angle YIX = 22^{\circ}\), then the angle \( \angle ZIY=74^{\circ}+ 22^{\circ}=96^{\circ}\)? No, that doesn't make sense. Wait, maybe the triangle is a right triangle? Wait, if \( ZN \) is perpendicular to \( YX \), then \( \angle ZNY = 90^{\circ}\). The incenter I, so \( \angle IZN=90^{\circ}-74^{\circ}=16^{\circ}\)? Wait, no, let's think again. The incenter is the intersection of angle bisectors. So \( \angle ZIX \) is the angle between the angle bisector of \( \angle Z \) and \( ZX \), and \( \angle YIX \) is the angle between the angle bisector of \( \angle Y \) and \( YX \). If \( ZN \) is perpendicular to \( YX \), then \( \angle ZNY = 90^{\circ}\). The angle \( \angle IZN \) is \( 90^{\circ}-\angle ZIX \). Given \( \angle ZIX = 74^{\circ}\), then \( \angle IZN=90 - 74=16^{\circ}\)? Wait, maybe the correct approach is: Since I is the incenter, \( \angle IZN = 90^{\circ}-\angle ZIX \). If \( \angle ZIX = 74^{\circ}\), then \( \angle IZN=16^{\circ}\). Wait, maybe I made a mistake. Alternatively, if we consider that the sum of angles in triangle \( IZN \): if \( \angle ZNI = 90^{\circ}\) (since \( ZN \perp YX \)), and \( \angle ZIN = 74^{\circ}\), then \( \angle IZN=180^{\circ}-90^{\circ}-74^{\circ}=16^{\circ}\).

Step1: Assume \( ZN\perp YX \), so \( \angle ZNY = 90^{\circ}\).

Step2: In triangle \( IZN \), we know that \( \angle ZNI = 90^{\circ}\) and \( \angle ZIN = 74^{\circ}\). Using the angle - sum property of a triangle (\( \angle IZN+\angle ZIN+\angle ZNI = 180^{\circ}\)), we have \( \angle IZN+74^{\circ}+90^{\circ}=180^{\circ}\).

Step3: Solve for \( \angle IZN \): \( \angle IZN=180^{\circ}-90^{\circ}-74^{\circ}=16^{\circ}\).

Answer:

\( HE = 7.5 \), \( BE = 22.5 \)

Problem 7: