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What must A measure in order for ABC to be isosceles?

Are you talking about the angle A. If you are then at what point of the triangle is the angle A.


Write a paragraph proof to show that the base angles of an isosceles trapezoid are congruent?

To prove that the base angles of an isosceles trapezoid are congruent, consider an isosceles trapezoid ( ABCD ) with ( AB \parallel CD ) and ( AD \cong BC ). By the properties of parallel lines, the angles ( \angle DAB ) and ( \angle ABC ) are consecutive interior angles formed by the transversal ( AD ) and ( BC ), respectively, thus ( \angle DAB + \angle ABC = 180^\circ ). Similarly, the angles ( \angle ADC ) and ( \angle BCD ) also sum to ( 180^\circ ). Since ( AD \cong BC ) and the trapezoid is isosceles, the two pairs of opposite angles must be equal, leading to ( \angle DAB \cong \angle ABC ) and ( \angle ADC \cong \angle BCD ), proving that the base angles ( \angle DAB ) and ( \angle ABC ) are congruent.


ABC is an isosceles right triangle AB has a slope of neg 1 and m ABC ninety degree's ABC is dilated by a sf of 1.8 with origin as center of dilation resulting in the image ABC what is slope of BC?

In an isosceles right triangle ( ABC ) with a right angle at ( B ) and ( AB ) having a slope of (-1), the slope of ( AC ) would be ( 1 ) since the two legs of the triangle are perpendicular. When the triangle is dilated with a scale factor of ( 1.8 ) from the origin, the slopes of the sides remain unchanged. Therefore, the slope of ( BC ), which is the leg opposite the right angle, remains ( 1 ).


If the median to a side of a triangle is also an altitude to that side then the triangle is isosceles How do you write this Proof?

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.


If an isosceles triangle has two equal sides that are 14 feet long and the third side is 17 feet long what are the angles?

Let the Isosceles Triangle be ∆ ABC with sides AB = AC = 14', and BC = 17' Draw a line bIsecting angle BAC. This line will be perpendicular to and bisect BC at point D. Then ∆ DBA (or ∆ DCA) is a right angled triangle with AB the hypotenuse. Angle ABD = Angle ABC is one of the two equal angles of the isosceles triangle. Cos ABD = BD/AB = 8.5/14 = 0.607143, therefore Angle ABC = 52.62° The third angle of the triangle is 180 - (2 x 52.62) = 180 - 105.24 = 74.76° The angles are therefore 52.62° , 52.62° and 74.76° .