ABCD is a parallelogram angle DAB=60. If the bisectors of angles A and B meet at P on CD then: (A)DP=1/3PC (B)PC=1/3DC (C)DP=1/3AB (D)DP=1/2AB
Answers
In the below figure, ABCD is a parallelogram and DAB = 60 . If the bisector AP and BP of angles A and B respectively meet P on CD. Prove that P is the midpoint of CD.
Since AB∥DC and Ap is a transversal, ∠PAB=∠DPA (Alternate angles) but,∠PAB=∠DAP (Given) ∠DPA=∠DAP=30º so, DP=DA (Isosceles triangle Property) similarly,PC=CB from(I) and (II), DP=P
Sol: ABCD is a parallelogram, in which ∠A = 60° ⇒ ∠B = 120° [Adjacent angles of a parallelogram are supplementary] ∠C = 60° = ∠A [Opposite angles of a parallelogram are equal] ∠D = ∠B = 120° [Opposite angles of parallelogram are equal] AP bisects ∠A ⇒ ∠DAP = ∠PAB = 30° BP bisects ∠B ⇒ ∠CBP = ∠PBA = 60° In ΔPAB, ∠APB = 90° [Angle sum property] In ΔPBC, ∠BPC = 60° [Angle sum property] In ΔADP, ∠APD = 30° [Angle sum property] In ΔPBC, ∠BPC = ∠CBP = 60° [Linear angles are supplementary] ⇒ BC = PC [Sides opposite to equal angles of a triangle are equal] -------- (1) In ΔADP, ∠APD = ∠DAP = 30° ⇒ AD = DP [Sides opposite to equal angles of a triangle are equal] But AD = BC [Opposite sides of parallelogram are equal] So, BC = DP -------- (2) From (1) and (2), we get DP = PC ⇒ P is the midpoint of CD.