# Chapter 4 Exercise 1

## 1a.

Given circle with center at O and points P,Q,R on the circle. Create angle POR and angle PQR. Construct a line through OQ and label the intersection of the line with the circle S. So QS is the diameter. Line segments OP, OQ, and OR are all radii and thus are congruent. Hence we have created two isosceles triangles, triangle ORQ and triangle OPQ. Since base angles of isosceles triangles are congruent, angle ORQ is congruent to angle OQR and angle OPQ is congruent to angle OQP. Note that angle SOR and angle SOP are exterior angles to triangle ORQ and triangle OPQ respectively. Note the measurement of angle SOR = the measurement of angle ORQ + measurement of angle OQR. = 2 times the measurement of angle OQR. The measurement of angle SOP = measurement of OPQ + measurement of angle OQP. = 2 time the measurement of OQP. Thus, the measurement of angle POR = the measurement of angle SOR + the measurement of angle SOP. = 2 times the measurement of angle OQR + 2 times measurement of OQP. = 2 time the measurement of angle OQR + angle OQP. = 2 times the measurement of PQR. In words, the measurement of the central angle POR is twice the measurement of the inscribed angle PQR.

## 1b.

Let O be the center of a circle with a diameter PR. Let Q be a point on the circle. the measure of angle POR=180, since PR is a diameter. Angle PQR is an inscribed angle. By exercise 1a, we know that the measurement of the central angle is twice the measurement of the inscribed angle. Thus the measurement of angle PQR is half the measurement of angle POR= 1/2 (180)=90. Therefore angle PQR is a right angle. Therefore any angle inscribed in a semicircle is a right angle.

## 1c.

Let ABC be triangle and let AM be the median from A to the side BC. Since M is the midpoint of BC, BM is congruent to CM. Construct the perpendicular line from A to side BC. Label the point intersection of the perpendicular line to BC D. Thus AD is the height of triangle ABC, triangle AMB, and triangle AMC. So the area of the triangles follows: area triangle = 1/2 (BC)(AD). Area triangle AMB= 1/2 (BM)(AD). Area triangle AMC= 1/2 (CM)(AD). Since M is the midpoint of BC, BM=CM=1/2 (BC). Hence, the area of triangle AMB= area triangle AMC= 1/2 area triangle ABC. Therefore, a median divides the triangle into two triangles of equal area.

## 1d.

Given triangle ABC with medians AX, BY, and CZ. Thus X,Y, and Z are midpoints of sides BC, CA, and AB. Recall, a median is a Cevian. We will use the converse of Ceva's Thm to prove the medians are concurrent. Since X,Y, and Z are midpoints of BC, Ca, and AB respectfully, we have: AZ=ZB, BX=XC, CY=YA. Thus, AZ/ZB=1, BX/XC=1, CY/YA= 1. Hence, AZ/ZB* BX/XC* CY/YA=1*1*1=1. Therefore, by the converse of CEva's Thm, the medians are concurrent.

## 1e.

Given triangle ABC with altitudes AX, BY, and CZ. Then AX, BY, and CZ are Cevians. To show AX, BY, and CZ are concurrent, show; AX/ZB*BX/XC*CY/YA=1. Angle BAY is congruent to angle CAZ because it is the same angle. Angle BYA is congruent to angle CZA because they are both right angles. Thus, triangle BAY is similar to triangle CAZ by AA similarity. Thus, AY/AZ=BY/CZ. Angle CBZ is congruent to angle ABX because they are the same angle. Angle BZC is congruent to angle AXB because they are both right angles. Thus, triangle CBZ is similar to triangle ABX by AA similarity. Thus, BX/BZ=AX/CZ. Angle BYC is congruent to angle AXC because they are both right angles. Angle BCY is congruent to angle ACX because they are the same angle. Thus, triangle BCY is similar to triangle ACX by AA similarity. Thus, CY/CX=BY/AX.

## 1h.

The opposite interior angles ofa convex cyclic quadrilateral and supplementary. Construct a circle with center O and convex cyclic quadrilateral ABCD. Without loss of generality, we will show that angle A and and C are supplementary. Construct a chord BD and radii OB and OD. In exercise 1a. we proved that the smaller of the angel BOD is twice the size of angle A. That is the measure of angel BOD = 2 times the measure of angle A. So measure of angle A = (measure of BOD)/2. Since the sum of the smaller angle BOD and the larger angle BOD is 360 degrees. Also by exercise 1A. We know that the larger angle BOD is twice the size of angle C. 360-measure BOD = 2 measure of C. So, measure of C = (360-measure of BOD)/2. Thus, the measure of A + measure of C= (Measure of BOD)/2 + (360-measure of BOD)/2. = 1/2 (measure of BOD + 360- measure of BOD). = 1/2 (360) = 180. Therefore angle A and angle C are supplementary