In an isosceles trapezoid the diagonals are congruent (true or false)
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Let ABCD be the trapezium with AB || CD, AB < DC and AD = BC. From B drop a perpendicular to meet CD at P. From A drop a perpendicular to meet CD at Q. In the triangles BPC, AQD: BP = AQ (opposite sides of rectangle ABPD) BC = AD (given) Angle BPC = angle AQD (both 90deg) Therefore triangles BPC and AQD are congruent (RHS), and in particular angle CBP = angle DAQ .....(1) Angle DAB = angle DAQ + angle QAB = angle DAQ + 90deg Angle CBA = angle CBP + angle PBA = angle CBP + 90deg Therefore from (1): Angle DAB = angle CBA .......(2) In the triangles ABC, BAD: AB = BA (common) BC = AD(given) Angle CBA = angle DAB (proved (2)) Therefore triangles ABC, BAD are congruent, and in particular AC = BD.
So True.
thank you sooooo muchhhh !!!!
true
true @angelacuevas
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