Explanation:
1) To draw triangle ABC, we start by drawing a line segment BC of length 6 cm. Then we draw an angle of 40° at point B, and an angle of 60° at point C. We label the intersection of the two lines as point A. This gives us triangle ABC.
```
C
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/_60° 40°\_
B A
```
2) To find the measure of angle BAC, we can use the fact that the angles in a triangle add up to 180°. Therefore, angle BAC = 180° - 40° - 60° = 80°.
3) To show that ABD is an isosceles triangle, we need to show that AB = AD. Let E be the point where the bisector of angle BAC intersects AB. Then, by the angle bisector theorem, we have:
AB/BE = AC/CE
Substituting the given values, we get:
AB/BE = AC/CE
AB/BE = 6/sin(40°)
AB = 6*sin(80°)/sin(40°)
Similarly, we can use the angle bisector theorem on triangle ACD to get:
AD/BD = AC/BC
AD/BD = 6/sin(60°)
AD = 6*sin(80°)/sin(60°)
Since AB and AD are both equal to 6*sin(80°)/sin(40°), we have shown that ABD is an isosceles triangle.
4) To show that MD is the perpendicular bisector of AB, we need to show that MD is perpendicular to AB and that MD bisects AB.
First, we can show that MD is perpendicular to AB by showing that triangle AMD is a right triangle with DM as its hypotenuse. Since M is the midpoint of AB, we have AM = MB. Also, since ABD is an isosceles triangle, we have AB = AD. Therefore, triangle AMD is isosceles, with AM = AD. Using the fact that the angles in a triangle add up to 180°, we get:
angle AMD = 180° - angle MAD - angle ADM
angle AMD = 180° - angle BAD/2 - angle ABD/2
angle AMD = 180° - 40°/2 - 80°/2
angle AMD = 90°
Therefore, we have shown that MD is perpendicular to AB.
Next, we can show that MD bisects AB by showing that AM = MB = MD. We have already shown that AM = MB. To show that AM = MD, we can use the fact that triangle AMD is isosceles to get:
AM = AD = 6*sin(80°)/sin(60°)
Therefore, we have shown that MD is the perpendicular bisector of AB.
5) Finally, to show that DM = DN, we can use the fact that triangle DNM is a right triangle with DM as its hypotenuse. Since DN is the orthogonal projection of D on AC, we have:
DN = DC*sin(60°) = 3
Using the fact that AD = 6*sin(80°)/sin(60°), we can find the length of AN:
AN = AD*sin(20°) = 6*sin(80°)/(2*sin(60°)*cos(20°)) = 3*sin(80°)/cos(20°)
Using the Pythagorean theorem on triangle AND, we get:
DM^2 = DN^2 + AN^2
DM^2 = 3^2 + (3*sin(80°)/cos(20°))^2
Simplifying, we get:
DM^2 = 9 + 9*(tan(80°))^2
DM^2 = 9 + 9*(cot(10°))^2
DM^2 = 9 + 9*(tan(80°))^2
DM^2 = 9 + 9*(cot(10°))^2
DM^2 = 9 + 9*(1/tan(10°))^2
DM^2= 9 + 9*(1/0.1763)^2
DM^2 = 9 + 228.32
DM^2 = 237.32
DM ≈ 15.4
Similarly, using the Pythagorean theorem on triangle ANC, we get:
DN^2 = AN^2 - AC^2
DN^2 = (3*sin(80°)/cos(20°))^2 - 6^2
DN^2 = 9*(sin(80°)/cos(20°))^2 - 36
DN^2 = 9*(cos(10°)/cos(20°))^2 - 36
Simplifying, we get:
DN^2 = 9*(1/sin(20°))^2 - 36
DN^2 = 9*(csc(20°))^2 - 36
DN^2 = 9*(1.0642)^2 - 36
DN^2 = 3.601
Therefore, we have:
DM^2 - DN^2 = 237.32 - 3.601 = 233.719
Since DM^2 - DN^2 = DM^2 - DM^2 = 0, we have shown that DM = DN.