Because ALL triangles total 180o...
Yes, a parallelogram or a rhombus would fit the given description.
It is a rhombus that fits the given description.
Hl & ha
A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel Let ABCD be a quadrilateral in which ABCD and AB=CD, where means parallel to. Construct line AC and create triangles ABC and ADC. Now, in triangles ABC and ADC, AB=CD (given) AC = AC (common side) Angle BAC=Angle ACD (corresponding parts of corresponding triangles or CPCTC) Triangle ABC is congruent to triangle CDA by Side Angle Side Angle BCA =Angle DAC by CPCTC And since these are alternate angles, ADBC. Thus in the quadrilateral ABCD, ABCD and ADBC. We conclude ABCD is a parallelogram. var content_characters_counter = '1032';
congruent
Because ALL triangles total 180o...
(1) vertical angles, (2) congruent triangles
Yes, a parallelogram or a rhombus would fit the given description.
Corresponding
It is a rhombus that fits the given description.
If two angles and the side opposite one of them in one triangle are equal to one side and two similarly located angles in a second triangle then the two triangles are congruent. (The triangles are exactly the same shape and size as each other).
Hl & ha
It doesn't imply they are congruent. However it doesn't mean they are not either. Not enough information has been given to establish their congruence.
A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel Let ABCD be a quadrilateral in which ABCD and AB=CD, where means parallel to. Construct line AC and create triangles ABC and ADC. Now, in triangles ABC and ADC, AB=CD (given) AC = AC (common side) Angle BAC=Angle ACD (corresponding parts of corresponding triangles or CPCTC) Triangle ABC is congruent to triangle CDA by Side Angle Side Angle BCA =Angle DAC by CPCTC And since these are alternate angles, ADBC. Thus in the quadrilateral ABCD, ABCD and ADBC. We conclude ABCD is a parallelogram. var content_characters_counter = '1032';
The HA and HL theorems for right triangles or the Pythagorean theorem might be of use.
A square would fit the given description.