Practice: special lines in a triangle

Geometry, grade 10 (4 units) Lesson 19 of 20Practice

Practice: special lines in a triangle

1

In the triangle $\displaystyle OAC$ : $\displaystyle O(0,0)$ , $\displaystyle A(15,0)$ and $\displaystyle C(33,24)$ .
The point $\displaystyle B(21,8)$ is on the side $\displaystyle AC$ .
Through $\displaystyle B$ goes a line parallel to the $\displaystyle x$ -axis, and it crosses the side $\displaystyle OC$ at the point $\displaystyle D$ .
The equation of the line $\displaystyle OC$ : $\displaystyle {y}={\frac{8}{11}x}$ .

\displaystyle y=\frac{8}{11}x

\displaystyle y=\frac{4}{3}x-20

\displaystyle y=8

\displaystyle O=(0,0)

\displaystyle A=(15,0)

\displaystyle B=(21,8)

\displaystyle C=(33,24)

\displaystyle D=(11,8)

a

Find the $\displaystyle x$ -coordinate of the point $\displaystyle D$ .

b

Show that $\displaystyle {DB}={AB}$ .

\displaystyle DB

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle AB

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

c

Prove that $\displaystyle AD$ bisects the angle $\displaystyle CAO$ .
(A hint: the triangle $\displaystyle ABD$ is isosceles, and $\displaystyle DB\parallel OA$ .)

d

Find $\displaystyle \angle BAO$ and $\displaystyle \angle BAD$ , rounded to two decimal places.
(A hint: the slope of the line $\displaystyle AC$ .)

\displaystyle \angle BAO

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle \angle BAD

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

2

The vertices of the triangle $\displaystyle ABC$ are $\displaystyle A(10,7)$ , $\displaystyle B(20,11)$ and $\displaystyle C(8,-5)$ .
$\displaystyle D$ is the midpoint of $\displaystyle AC$ and $\displaystyle E$ the midpoint of $\displaystyle AB$ .
The lines $\displaystyle CE$ and $\displaystyle BD$ meet at the point $\displaystyle M$ .

a

\displaystyle x_M

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle y_M

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

The ratio

\displaystyle \frac{EM}{EC}

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

3

The triangle $\displaystyle ABC$ is isosceles: $\displaystyle {AB}={AC}$ .
$\displaystyle BD$ and $\displaystyle CE$ are the medians to the legs.
Prove: $\displaystyle {BD}={CE}$ .

4

In the triangle $\displaystyle ABC$ , $\displaystyle BD$ is the altitude to the side $\displaystyle AC$ , and $\displaystyle CE$ the altitude to the side $\displaystyle AB$ .
$\displaystyle {BD}={CE}$ .
Prove: $\displaystyle {AB}={AC}$ .
(A hint: the triangles $\displaystyle BEC$ and $\displaystyle CDB$ .)

5

In the triangle $\displaystyle ABC$ , $\displaystyle {AB}={10}$ , $\displaystyle {AC}={15}$ and $\displaystyle {BC}={20}$ .
$\displaystyle AD$ bisects the angle $\displaystyle A$ .
Find $\displaystyle BD$ .

6

$\displaystyle A(2,1)$ and $\displaystyle B(6,5)$ .

a

Write on the board the equation of the perpendicular bisector of the segment $\displaystyle AB$ .

\displaystyle A=(2,1)

\displaystyle B=(6,5)

b

Find a point on the $\displaystyle y$ -axis at equal distances from $\displaystyle A$ and from $\displaystyle B$ .
Explain why it is.

Formulas: geometry, grade 10 (4 units)

Analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
A line’s equation in explicit form – $\displaystyle m$ the slope, $\displaystyle b$ where it meets the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
A line’s equation in implicit form
$\displaystyle Ax+By+C=0$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
The slope – $\displaystyle \alpha$ the angle between the line and the positive direction of the $\displaystyle x$ -axis
$\displaystyle m$ $\displaystyle =$ $\displaystyle \tan\alpha$

Similar triangles and midsegments

The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
The ratio of the perimeters
$\displaystyle \frac{P_1}{P_2}$ $\displaystyle =$ $\displaystyle k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$
The ratio of corresponding altitudes, medians and angle bisectors
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{d_1}{d_2}=k$
A triangle’s midsegment – $\displaystyle DE$ parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$

Trigonometry

Sine – $\displaystyle a$ the side opposite the angle, $\displaystyle c$ the hypotenuse
$\displaystyle \sin\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{c}$
Cosine – $\displaystyle b$ the side adjacent to the angle
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{b}{c}$
Tangent – the opposite side over the adjacent side
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{b}$
The tangent from the sine and the cosine
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{\sin\alpha}{\cos\alpha}$
The identity from Pythagoras
$\displaystyle \sin^{2}\alpha+\cos^{2}\alpha$ $\displaystyle =$ $\displaystyle 1$
Angles whose sum is $\displaystyle 90^{\circ}$
$\displaystyle \sin(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \cos\alpha$
$\displaystyle \cos(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The sine of an obtuse angle
$\displaystyle \sin(180^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
A triangle’s area – $\displaystyle \gamma$ the angle between the sides $\displaystyle a$ and $\displaystyle b$
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}ab\sin\gamma$

Exact values

$\displaystyle \sin30^{\circ}=\cos60^{\circ}=\frac{1}{2}$
$\displaystyle \sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}$
$\displaystyle \sin60^{\circ}=\cos30^{\circ}=\frac{\sqrt{3}}{2}$
$\displaystyle \tan30^{\circ}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{3}}{3}$
$\displaystyle \tan45^{\circ}$ $\displaystyle =$ $\displaystyle 1$
$\displaystyle \tan60^{\circ}$ $\displaystyle =$ $\displaystyle \sqrt{3}$

Special lines in a triangle

Where the medians meet, $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Where the medians meet, in coordinates
$\displaystyle G\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)$
The median $\displaystyle AD$ splits the triangle into two triangles of equal area
$\displaystyle S_{ABD}$ $\displaystyle =$ $\displaystyle S_{ACD}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$