0(zero)
Angle of incidence = angle of reflection
The Angle Bisector Theorem states that given triangle and angle bisector AD, where D is on side BC, then . Likewise, the converse is also true. Not sure if this is what you want?
To prove that the base angles of an isosceles trapezoid are congruent, consider an isosceles trapezoid ( ABCD ) with ( AB \parallel CD ) and ( AD \cong BC ). By the properties of parallel lines, the angles ( \angle DAB ) and ( \angle ABC ) are consecutive interior angles formed by the transversal ( AD ) and ( BC ), respectively, thus ( \angle DAB + \angle ABC = 180^\circ ). Similarly, the angles ( \angle ADC ) and ( \angle BCD ) also sum to ( 180^\circ ). Since ( AD \cong BC ) and the trapezoid is isosceles, the two pairs of opposite angles must be equal, leading to ( \angle DAB \cong \angle ABC ) and ( \angle ADC \cong \angle BCD ), proving that the base angles ( \angle DAB ) and ( \angle ABC ) are congruent.
Label the triangle ABC. Draw the bisector of angle A to meet BC at D. Then in triangles ABD and ACD, angle ABD = angle ACD (equiangular triangle) angle BAD = angle CAD (AD is angle bisector) so angle ADB = angle ACD (third angle of triangles). Also AD is common. So, by ASA, triangle ABD is congruent to triangle ACD and therefore AB = AC. By drawing the bisector of angle B, it can be shown that AB = BC. Therefore, AB = BC = AC ie the triangle is equilateral.
It's infinitely small. It can't be defined. Any angle can be divided ad infinitum. The best we can say is that it's close to zero.
8 sides
Specular reflection is the mirror-like reflection of light(or of other kinds of wave) from a surface, in which light from a single incoming direction (a ray) is reflected into a single outgoing direction. Such behavior is described by the law of reflection, which states that the direction of incoming light (the incident ray), and the direction of outgoing light reflected (the reflected ray) make the same angle with respect to the surface normal, thus the angle of incidence equals the angle of reflection ( in the figure), and that the incident, normal, and reflected directions are coplanar. This behavior was first discovered through careful observation and measurement by Hero of Alexandria (AD c. 10-70).
1960 years
Ad is higher than ac
143
That is not necessarily true.
Given: AD perpendicular to BC; angle BAD congruent to CAD Prove: ABC is isosceles Plan: Principle a.s.a Proof: 1. angle BAD congruent to angle CAD (given) 2. Since AD is perpendicular to BC, then the angle BDA is congruent to the angle CDA (all right angles are congruent). 3. AD is congruent to AD (reflexive property) 4. triangle BAD congruent to triangle CAD (principle a.s.a) 5. AB is congruent to AC (corresponding parts of congruent triangles are congruent) 6. triangle ABC is isosceles (it has two congruent sides)
If AC = 10 units and D is the midpoint of AC then AD = AC/2 = 5 units!
The Angle Bisector Theorem states that given triangle and angle bisector AD, where D is on side BC, then . Likewise, the converse is also true. Not sure if this is what you want?
AC is alternating current and AD is Anno Domini or after Christ.
That is not necessarily true.
what is that...
Label the triangle ABC. Draw the bisector of angle A to meet BC at D. Then in triangles ABD and ACD, angle ABD = angle ACD (equiangular triangle) angle BAD = angle CAD (AD is angle bisector) so angle ADB = angle ACD (third angle of triangles). Also AD is common. So, by ASA, triangle ABD is congruent to triangle ACD and therefore AB = AC. By drawing the bisector of angle B, it can be shown that AB = BC. Therefore, AB = BC = AC ie the triangle is equilateral.