You cannot prove that because it's false
Given:In a triangle ABC in which EF BC To prove that:AE/EB=AF/FC Construction:Draw EX perpendicular AC and FY perpendicular AB Proof:taking the ratios of area of triangle AEF and EBF and second pair of ratio of area of triangle AEF and ECF. We get AE/EB and AF/FC we know that triangle lie b/w sme and same base is equal in area therefore area of EBF I equal to area of ECF therefore AE/EB=AF/FC HENCE PROVED
Sweet hilarity! Napoleon Dynamite is out to prove to one and all that he's got nothing to prove. How dramatic!
proved that America was number one in technology and therefore the number one superpower
Pythagorean's Theorem is one of the most famous ones. It says that the two squared sides of a right triangle equal the squared side of the hypotenuse. In other words, a2 + b2 = c2
This one here:( n / 3 ) ^ n < n ! < ( n / 2 ) ^ n
No, but there is a way to prove that zero equals one.
One plus one does equal two. If you have one object, then bring in another object, you will have one more than what you started with. That would be two.
you can prove any one of these statements to prove that quadrilateral is a rectangle: -- Opposite sides are parallel and any one angle is a right angle. -- Opposite sides are equal and any one angle is a right angle. -- All four angles are right angles. -- Adjacent angles are complementary, and one of them is a right angle. -- Opposite sides are either equal or parallel, and area is equal to the product of two adjacent sides. -- Diagonals are equal.
You prove that the two sides (not the bases) are equal in length. Or that the base angles are equal measure.
prove any two adjacent triangles as congruent
You need to prove that there are two pairs of sides with equal lengths. Imagine a rectangle A C B D Since side AB is equal in length to side CD that is one equal pair. Likewise, AC is equal to BD and is the other equal pair. Since there are two equal parallel pairs, the angles must all be 90 degrees by rules of geometry so it is a rectangle. Note: The same thing may be used to prove a square, except a square has all sides of the same length. ================================================ If you can prove any one of these statements, then your quadrilateral is a rectangle: -- Opposite sides are parallel and any one angle is a right angle. -- Opposite sides are equal and any one angle is a right angle. -- All four angles are right angles. -- Adjacent angles are complementary, and one of them is a right angle. -- Opposite sides are either equal or parallel, and area is equal to the product of two adjacent sides. -- Diagonals are equal.
YES ... to prove it, divide both the numerator and the denominator of 5/15 by 5 and you will get 1/3
It isn't equal, and any proof that they are equal is flawed.
a parallelogram with opposite equal acute angles, opposite equal obtuse angles, and four equal sides.
It is not true so you cannot prove it. You can concoct a "proof" that might look OK but it will be flawed.
to prove cash you look at the amount of money you have and accounting books. if the value is equal then you have proved cash
to prove cash you look at the amount of money you have and accounting books. if the value is equal then you have proved cash