In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles.

Solution

ABCD is a parallelogram. The angle bisectors AE and BE of adjacent angles A and B meet at E.

To Prove: ∠AEB = 90°
Proof: ∵ AD || BC
| Opposite sides of ||gm and transversal AB intersects them
∴ ∠DAB + ∠CBA = 180°
| ∵ Sum of consecutive interior angles on the same side of a transversal is 180°
⇒ 2∠EAB + 2∠EBA = 180°
| ∵ AE and BE are the bisectors of ∠DAB and ∠CBA respectively.
⇒ ∠EAB + ∠EBA = 90° ...(1)
In ∆EAB,
∠EAB + ∠EBA + ∠AEB = 180°
| ∵ The sum of the three angles of a triangle is 180°
⇒ 90° + ∠AEB = 180° | From (1)
⇒ ∠AEB = 90°.

Some More Questions From Circles Chapter

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that:

(i) ∆APD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB ≅ ∆CPD
(iv) AQ = CP
(v) APCQ is a parallelogram.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively (see figure). Show that:
(i) ∆APB ≅ ∆CQD
(ii) AP = CQ.

In ∆ABC and ∆DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, Band C are joined to vertices D, E and F respectively (see figure). Show that:
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF
(vi) ∆ABC ≅ ∆DEF. [CBSE 2012

ABCD is a trapezium in which AB || CD and AD = BC (see figure): Show that
(i) ∠A = ∠B
(ii)    ∠C = ∠D
(iii)    ∆ABC = ∆BAD
(iv)    diagonal AC = diagonal BD.

[Hint. Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]