Let the triangle be ABC and suppose the median AD is also an altitude.AD is a median, therefore BD = CD
AD is an altitude, therefore angle ADB = angle ADC = 90 degrees
Then, in triangles ABD and ACD,
AD is common,
angle ADB = angle ADC
and BD = CD
Therefore the two triangles are congruent (SAS).
And therefore AB = AC, that is, the triangle is isosceles.
45
You could try the fact that the altitude of an equilateral traingle is also its median so that it divides the base in half. Then Pythagoras does the rest.
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
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You could try the fact that the altitude of an equilateral traingle is also its median so that it divides the base in half. Then Pythagoras does the rest.
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.
The following is the answer.
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.
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.
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)
There cannot be a proof because the statement need not be true.
how the hell do you even find the centroid of a triangle to begin with, that's what i want to know!
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.