ABCD is a parallelogram in which angle A = 60 degree, if the bisectors of angle A and angle B meet DC at P, prove that (i) angle APB = 90 degree (ii) AD = DP and PB = PC=BC (iii) DC = 2AD
Answers
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.
Given: ABCD is a parallelogram in which ∠A = 60°. AP and PB are the bisector of ∠A and ∠B respectively.
∠PAB = ∠PAD = 30° (AP bisects ∠A)
and ∠ABP = ∠PBC = 60° (BP bisects ∠B)
i)
In triangle APB, using angle sum property of triangle, we get
∠APB + ∠PAB + ∠ABP = 180°
⇒ ∠APB + 30° + 60° = 180°
⇒ ∠APB = 90°
ii)
AB || CD and AP is the transversal.
∴ ∠PAB = ∠APD = 30° (Alternate angles)
In ΔAPD,
∠PAD = ∠APD = 30°
∴ PD = AD ...(1) (Equal sides have equal angles opposite to them)
AD || BC and AB is the transversal.
∴ ∠A + ∠B = 180° (Sum of adjacent interior angles is 180°)
⇒ 60° + ∠B = 180°
⇒ ∠B = 120°
⇒ ∠PBA = ∠PBC = 60° (PB is bisector ∠B)
∠PBA = ∠BPC = 60° (Alternate angles)
In ΔPBC,
∠BPC = ∠PBC = 60°
Also, ∠PCB = 60° [Using angle sum property of triangle]
So, ΔPBC is an equilateral triangle
∴ BC = PB = PC ...(2) (Equal sides of an equilateral triangle)
iii)
CD = DP + PC
∴ CD = AD + BC [Using (1) and (2)]
⇒ CD = AD + AD [AD = BC (Opposite sides of parallelogram are equal)]
⇒ CD = 2AD