Question

Prove that “ the quadrilateral formed by the intersection of all the interior angle on the same side of the transversal is a rectangle”

PLS READ THE QUESTION CAREFULLY

Answer 1

Answer:

IN SHORT

since, ABCD is a parallelogram

then, DC ll AB

and DA is the transversal

thus, PQRS is a quadrilateral

whose angle each angle 90°

PQRS is a rectangle

IN EXPLANATION

Given,

Let ABCD be a parallelogram

To prove quadrilateral PQRS is a triangle

since, ABCD is a parallelogram, then DC ll AB and DA is the transversal

angle A + angle D = 180° [ sum of conterior angles of a parallelogram is 180° ]

1/2 angle A + 1/2 angle D = 90° [ dividing both sides by 2 ]

angle SAD + angle SDA = 90°

angle ASD = 90° [ since, sum of all angles of a triangle is 180° ]

... PQRS is a quadrilateral whose each angle is 90°

Hence, PQRS is a rectangle

Hope it helps you

Answer 2

Answer:

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