Question

$ABCD$ is a parallelogram and $\angle DAB={60}^o$. If the bisects $AP$ and $BP$ of angles $A$ and $B$ respectively, meet at $P$ on $CD$.Prove that $P$ is the midpoint of $CD$.

Answer 1

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