In a parallelogram show that bisectors of adjacent angles intersect at right angles?

Answer 1

Label the parallelogram ABCD (with AB parallel to DC, AD parallel to BC) and the point where the angle bisectors of DAB and ABC meet X. It is needed to be shown that angle AXB is 90o.

Call angle DAB a and angle ABC b. Then the bisectors of these angles are 1/2a and 1/2b.

First note that for a parallelogram, opposite angles are equal, ie:

ADC = ABC = a

DAB = DCB = b

Also, the sum of all angles in a quadrilateral (and a parallelogram is one) in 360o. Thus:

360o = DAB + ABC + BCD + CDA

360o = a +b + a + b

360o = 2a +2b

Or dividing by 2:

180o = a + b

and again by 2:

90o = 1/2a + 1/2b - [1]

Now consider triangle AXB. Angle XAB = 1/2b, XBA = 1/2b. So angle:

AXB = 180o - (XAB + XBA)

AXB = 180o - (1/2a + 1/2b)

AXB = 180o - 90o (As shown in [1] above)

AXB = 90o

Thus the bisectors of adjacent angles of a parallelogram meet at right angles.

Opposite angles of a parallelogram can be shown to be equal by extending sides AB and DC and then using corresponding and alternate angles.

Alternatively, adjacent angles of a parallelogram can also be shown to sum to 180o by extending line AB (to any point Y beyond B) and using that DAY and CBY are corresponding angles and that ABC and CBY sum to 180o since ABY is a straight line.