Mathematics
In the diagram given below, the bisectors of interior angles of the parallelogram PQRS enclose a quadrilateral ABCD. Show that:
(i) ∠PSB + ∠SPB = 90°
(ii) ∠PBS = 90°
(iii) ∠ABC = 90°
(iv) ∠ADC = 90°
(v) ∠A = 90°
(vi) ABCD is a rectangle
Related Questions
One of the diagonals of a rhombus and its sides are equal. Find the angles of the rhombus.
In a parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. Prove that :
(i) AE = AD
(ii) DE bisects ∠ADC and
(iii) Angle DEC is a right angle.
In a parallelogram ABCD, X and Y are mid-points of opposite sides AB and DC respectively. Prove that:
(i) AX = YC.
(ii) AX is parallel to YC
(iii) AXCY is a parallelogram.
The given figure shows a parallelogram ABCD. Points M and N lie in diagonal BD such that DM = BN. Prove that:
(i) △DMC ≅ △BNA and so CM = AN.
(ii) △AMD ≅ △CNB and so AM = CN.
(iii) ANCM is a parallelogram.
