MATH 5200 homework 9

Student solutions

1. The formulas refer to the triangle ABC, with sides a = BC, b = AC, c = AB. Following the usual shorthand notation, angles will be labelled by
their vertices.

I. Area formula:

Area = 1/2 ab sin(C)

Solution (Kristen Wagner):

GSP sketch:

In order to prove that the area of ABC is equal to (1/2) ab sin C, first drop the line through B that is perpendicular to the line that runs through AC.
A generalization of how this will be proved is to show that the area of ABC is equal to the area of ADB - the area of CDB.

Proof:

The area of triangle any triangle is (1/2)*b*h, where b is the base and h is the height. So, the

area of triangle ABD = (1/2)*b*h

= (1/2)(b + x)(y)

= (1/2)by + (1/2)xy

area of triangle BCD = (1/2)xy.

So the area of triangle ABC can be found by subtracting the area of BCD from the area of ABD. So, the

area ABC = area ABD - Area BCD

= (1/2)by + (1/2)xy -(1/2)xy

Now look at the sine function of the two right triangles.

sin (<DAB) = y/c

sin (<BCD) = y/a

Therefore, a*sin (<BCD) = y. The sin (<BCD) = sin (BCA) because supplementary angles have the same sine measure. So, the

area of ABC = (1/2)*b*h

= (1/2)*b*y

= (1/2)*b*a*sin (<BCA)

= (1/2) ab sin C.

Application - Area of a Quadrilateral inscribed in a circle

GSP sketch:

From the new formula for the area of a triangle, I was able to find a formula for the area of a quadrilateral inscribed in a circle. The formula for the
area is (using the figure above):

Area ABCD = (1/2) (sin C) (ab + ef).

This formula is derived by splitting the quadrilateral into two triangle. The area of triangle ACB is equal to (1/2) ab sin C and the area of triangle
ABD is equal to (1/2) ef sin D. Adding these two areas will give the area of the quadrilateral. But the

sin C = sin D because the two angles are supplementary. This is a property of inscribed quadrilaterals. So, the

area of ABCD = (1/2) ab sin C + (1/2) ef sin C

= (1/2)(sin C)(ab +ef).

II. Law of cosines:

c2 = a2 + b2 - 2ab cos(C).

Solution (Julie Rawlin):

As you see in the triangle above, I have constructed the altitude from B to AC, so now I have two right triangles. I can use the Pythagorean Theorem
now to help prove the law of cosines. First, I will let the green segment above, (DC) be x and the segment (AD) be b-x. Note that x + (b-x) = b. So if
we use the theorem, we have from triangle BDC that,

c2 = (b-x)2 + h2, where h is the altitude. Then we simplify and get

c2 = b2-2bx+x2+h2

c2 = b2-2bx+a2, now we can write x in terms of a and angle C, so

c2 = b2 &endash;2b(a cos (C)) +a2, and when we rewrite this we have,

c2 = a2 + b2 &endash;2ab cos (C). Thus we have the law of cosines.

III. Law of sines:

a/sin(A) = b/sin(B) = c/sin(C).

Solution (Nami Youn):

GSP sketch:

Proof of law of sines:

We have three cases; A triangle ABC is an acute, obtuse or right triangle.

Case I. Acute triangle

Let the center of circumscribed circle be O. Extend the line segment BO. Let the intersection point between this ray BO and the circle be A'.

Since the line segment BA' is the diameter of the circle, BA' = 2R.

Since angle A and angle A' are inscribed angles of the same arc BC, A = A' .

So, sinA = sinA'.

We proved that a(A'CB) = 90 when A'B is a diameter.

By definition, sinA = sinA' = BC/BA' = a/ 2R.

Then, a/sinA = 2R

Similary, b/sinB = c/sinC = 2R.

We proved a/sinA = b/sinB = c/sinC = 2R.

Case II. Obtuse triangle

We proved that opposite angles of a quadrilateral are supplementary if and only if the quadrilateral can be inscribed in a circle. So, A + A' = 180,
then A = 180 - A'.

Since a(A'CB) = 90,

By definition, sin A = sin(180-A') = sin A' = a/ 2R.

Then, a/sinA = 2R

Similary, b/sinB = c/sinC = 2R.

We proved a/sinA = b/sinB = c/sinC = 2R.

Case III. Right triangle

Since A = 90, sinA = 1. a = BC is the diameter of circle, so, a= 2R.

Therefore, a/ sinA = 2R.

Similary, b/sinB = c/sinC = 2R.

We proved a/sinA = b/sinB = c/sinC = 2R.

2. Prove that for a triangle ABC labelled as in problem 1, with circumradius R, the area of ABC is equal to abc/4R.

Solution (Whitney Burton):

Proof:

There are actually two different cases of this proof. There is the case when the triangle is obtuse and the case when the triangle is acute.

Case 1: Obtuse Triangle

- We need to show that the area of triangle ABC =abc/4R

- Construct the segment AH which is the diameter of the circle. Because it is the diameter of the circle we know that its length is 2R.

- Triangle ABC and triangle ABH subtend the the same chord AB, so angle ACB and agle AHB are supplementary. We know this because these
two triangles create a quadrilateral which is inscribed in a circle, which tells us that opposite angles are supplementary.

- We know that sin H = sin(180-C) and sin H = c/2R so, through simple algebra, the sin(180-C) = c/2R.

- Since the sin (180-C) = sin C, by negative congruence, then sin C = c/2R.

- The area of ABC = 1/2 ab sin(C) since sin (C) = c/2R = 1/2 ab(c/2R) = abc/4R.

Case 2: Acute Triangle

- When the triangle ABC is an acute triangle....

- Triangle ABC and triangle AHB, again, subtend the same chord AB.

- Therefore, angle ACB = angle AHB

- sin H = sin C

- sin H = c/2R = sin C

- The area of ABC = 1/2 absin C

- since sin C = c/2R, then the same algebra argument as above gives us . . . .

area of ABC = abc/4R.

3. Let ABC be a triangle with circumradius R. Suppose that C1 and C2 are circles such that both C1 and C2 pass through the point A, C1 is tangent
to BC at B, and C2 is tangent to BC at C. Let p be the radius of C1 and let q be the radius of C2. Prove that pq = R2.

Solution (Amanda Avery):

GSP sketch:

Proof:

Construct the unique circumcircle of triangle ABC. Let O be the center. Create the radii OA, OB, OC. So, OA = OB = OC = R. So, the triangles
AOC, BOA, and COB are isosceles triangles. And therefore, by the isosceles triangle theorem, we know that a(CAO) = a(OCA) = g; a(OAB) =
a(ABO) = e; and a(BCO) = a(OBC) = h. So, from the triangle angle sum theorem, we know that 2g + 2e + 2h = 180?, i.e. g + e + h = 90?. Now,
since c1 (with center D) passes through point A and is tangent to BC at B, then A and B are points on the circle and hence AD = BD = p, the
radius of c1. So DADB is isosceles, and by the isosceles triangle theorem, we know that a(BAD) = a(DBA) = d. Likewise, for c2 (with center E)
passes through point A and is tangent to BC at C, then A and C are points on the circle and hence AE = CE = q, the radius of c2. So, DAEC is
isosceles, and by the isosceles triangle theorem, we know that a(EAC) = a(ACE) = f. Now, consider that since c1 is tangent to BC at B, then the
angle that radius p makes with side BC is 90?. By the addition angle axiom, then we have a(DBC) = a(DBA) + a(ABO) + a(OBC) = d + e + h =
90?. So, from above we have g + e + h = 90?, but = d + e + h = 90?. So, we can conclude that g = d. So, by AA Similarity we have DDBA ~
DOCA. Now, a very similar argument is made for the similarity of DABO ~ DACE. So, we can set up the following ratios from these similar
triangles: p/R = c/b and R/q = c/b. Now, by transitivity, we have p/R = R/q, i.e. pq = R2.

4. Prove that for a triangle ABC labelled as in problem 1, if D is the midpoint of side AB, then the length m of the median CD satisfies the equation

m2 = 1/2 a2 + 1/2 b2 - 1/4 c2.

Solution (Marissa Henderson):

Law of cosines: c2 = a2 + b2 - 2ab cos(C)

From the law of cosines, we can find m2 as well as b2 and then solve for cos B and substitute cos B into our formula for m2.....

m2 = (1/2c)2 + a2 - 2(a)(1/2c) cos B

= 1/4c2 +a2 - ac cos B

and b2 = c2 + a2 - 2 ac cos B

if rearranged... (b2-c2-a2) / -2ac = cos B

when this ratio is substituted in for cos B of the m2 formula you get...

m2 = 1/4c2 +a2 - ac ((b2-c2-a2) / -2ac)

= 1/4c2 + a2 + 1/2b2 - 1/2c2 - 1/2a2

= 1/2a2 + 1/2b2 - 1/4c2