Augustus Pythagoras: let the equal sides be 1 unit.
The square of the third side, which is the hypotenuse, is equal to the sum of the squares of the other two sides, in this case 12 and 12, a total of 2. The hypotenuse is therefore equal to the square root of two.
45
Let the triangle be ABC and suppose the median AD is also an altitude.AD is a median, therefore BD = CDAD is an altitude, therefore angle ADB = angle ADC = 90 degreesThen, in triangles ABD and ACD,AD is common,angle ADB = angle ADCand BD = CDTherefore the two triangles are congruent (SAS).And therefore AB = AC, that is, the triangle is isosceles.
180 Proof: 30-60-90 Triangle 45-45-90 * * * * * The answer is correct, but two examples (or even a million) do not constitute mathematical PROOF.
The proof would finish with the statement:"Therefore, bc is congruent to de".
The isosceles triangle theorem states: If two sides of a triangle are congruent, then the angles opposite to them are congruent Here is the proof: Draw triangle ABC with side AB congruent to side BC so the triangle is isosceles. Want to prove angle BAC is congruent to angle BCA Now draw an angle bisector of angle ABC that inersects side AC at a point P. ABP is congruent to CPB because ray BP is a bisector of angle ABC Now we know side BP is congruent to side BP. So we have side AB congruent to BC and side BP congruent to BP and the angles between them are ABP and CBP and those are congruent as well so we use SAS (side angle side) Now angle BAC and BCA are corresponding angles of congruent triangles to they are congruent and we are done! QED. Another proof: The area of a triangle is equal to 1/2*a*b*sin(C), where a and b are lengths of adjacent sides, and C is the angle between the two sides. Suppose we have a triangle ABC, where the lengths of the sides AB and AC are equal. Then the area of ABC = 1/2*AB*BC*sin(B) = 1/2*AC*CB*sin(C). Canceling, we have sin(B) = sin(C). Since the angles of a triangle sum to 180 degrees, B and C are both acute. Therefore, angle B is congruent to angle C. Altering the proof slightly gives us the converse to the above theorem, namely that if a triangle has two congruent angles, then the sides opposite to them are congruent as well.
converse of the isosceles triangle theorem
converse of the isosceles triangle theorem
There cannot be a proof because the statement need not be true.
If vertices are at (7, 3) (4, -3) and (10, -3) then it is an isosceles triangle because by using the distance formula it has 2 equal sides of 3 times square root 5 and a 3rd side of 6.
45
Let the triangle be ABC and suppose the median AD is also an altitude.AD is a median, therefore BD = CDAD is an altitude, therefore angle ADB = angle ADC = 90 degreesThen, in triangles ABD and ACD,AD is common,angle ADB = angle ADCand BD = CDTherefore the two triangles are congruent (SAS).And therefore AB = AC, that is, the triangle is isosceles.
The smallest angle would be = 38 degrees. Proof: Base angles of an isosceles triangle must equ All angles of the triangle must add up to 180 degress considering that the known angle is not under 89 degrees the other two must equal, yet both add up to 76 degrees.
It can be shown that for any right angle triangle that its hypotenuse when square is equal to the sum of its squared sides.
Pythagoras most famous proof is the pythagorean proof . It states that in a right angled triangle , the square of hypoteneus ( the longest side of the triangle ) is equal to the sum of squares of the other two sides .
The following is the answer.
When calculating the area of a triangle, you need to divide by 2 because the formula for the area of a triangle is 1/2 multiplied by the base multiplied by the height. This is because a triangle is essentially half of a rectangle, and the formula reflects this relationship. Dividing by 2 ensures that you are finding the correct area for the given triangle based on its base and height dimensions.
The largest possible triangle is an equilateral triangle. Here's a sort of proof - try making some sketches to get the idea. * For any given isosceles triangle ABC that you might inscribe, where AB = BC... * ...Moving vertex A to be perpendicularly above the midpoint of BC will increase the area, since its distance from BC (the height of the triangle) will be at a maximum.* This gives a new isosceles, where AB = AC. * The same thing applies to the new isosceles. You can keep increasing the area in this way until the process makes no difference. If the process can increase the area no further, it can only be because all the vertices are already above the midpoints of the opposite edges. Which means we have an equilateral triangle. Anyhow, to answer the question, an equilateral triangle inscribed in a circle of radius r will have side length d where d2 = 2r2 - 2r2cos(120) from the cosine rule. and since cos(120) = -1/2 d2 = 2r2 + r2 = 3r2 and so d = r sqrt(3) *Equally, move vertex C above the midpoint of AB.