Show that the bisectors of angles of a parallelogram forms a rectangle.
Answers
⏩ 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....:-)
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