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How do you prove the isosceles triangle theorem? - Answers

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.



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How do you prove the isosceles triangle theorem? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_prove_the_isosceles_triangle_theorem

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.



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https://math.answers.com/math-and-arithmetic/How_do_you_prove_the_isosceles_triangle_theorem

How do you prove the isosceles triangle theorem? - Answers

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.

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      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.
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