Question
In Figure, ABCD is a parallelogram in which $\angle\text{A}=60^\circ$. If the bisectors of $\angle\text{A}$, and $\angle\text{B}$ meet at P, prove that AD = DP, PC = BC and DC = 2AD.
✓
Answer
AP bisects $\angle\text{A}$ Then, $\angle\text{DAP} = \angle\text{PAB} = 30^\circ$ Adjacent angles are supplementary Then, $\angle\text{A}+\angle\text{B}=180^\circ$$\angle\text{B}+60^\circ=180^\circ$
$\angle\text{B}=180^\circ-60^\circ$
$\angle\text{B}=120^\circ$
BP bisects $\angle\text{B}$ Then, $\angle\text{PBA} = \angle\text{PBC} = 30^\circ$$\angle\text{PAB} = \angle\text{APD} = 30^\circ$ [Alternate interior angles]
Therefore, AD = DP [Sides opposite to equal angles are in equal length] Similarly,$\angle\text{PAB} = \angle\text{APD} = 60^\circ$ [Alternate interior angles]
Therefore, PC = BC DC = DP + PC DC = AD + BC [Since, DP = AD and PC = BC] DC = 2AD [Since, AD = BC, opposite sides of a parallelogram are equal]
$\angle\text{B}=180^\circ-60^\circ$
$\angle\text{B}=120^\circ$
BP bisects $\angle\text{B}$ Then, $\angle\text{PBA} = \angle\text{PBC} = 30^\circ$$\angle\text{PAB} = \angle\text{APD} = 30^\circ$ [Alternate interior angles]
Therefore, AD = DP [Sides opposite to equal angles are in equal length] Similarly,$\angle\text{PAB} = \angle\text{APD} = 60^\circ$ [Alternate interior angles]
Therefore, PC = BC DC = DP + PC DC = AD + BC [Since, DP = AD and PC = BC] DC = 2AD [Since, AD = BC, opposite sides of a parallelogram are equal]
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