Math, asked by BiteGlory4, 1 year ago

Show that the bisectors of angles of a parallelogram forms a rectangle.​

Answers

Answered by Anonymous
48

\huge\mathfrak{Bonjour!!}

\huge\bold\purple{Answer:-}

Let P,Q,R and S be the points of intersection of the bisectors of angles A and B, B and C, C and D & D and A respectively of parallelogram ABCD.

Now,

Consider ASD,

Since, DS bisects angle D and AS bisects angle A,

Therefore,

angle DAS + angle ADS = 1/2 angle A + 1/2 angle D

= 1/2 (angle A + angle D)

=1/2 × 180° (angle A and angle D are interior angles on the same side of the transversal)

Also,

angle DAS + angle ADS + angle DSA = 180° (Angle sum property of triangle)

=> 90° + angle DSA = 180°

=> angle DSA = 90°

Therefore,

angle PSR = 90° (V.O.A)

|||ly , it can be shown that angle APB = 90° or angle SOQ = 90°

(As it was shown for angle DSA)

|||ly, angle PQR = 90° and angle SRQ = 90°

Therefore,

PQRS is a quadrilateral in which all angles are right angles and thus this shows that PQRS is a rectangle.

HENCE PROVED!

Hope it helps....:-)

Be Brainly...

WALKER

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