In the adjoining figure, ABCD is a parallelogram in which =60^ . If the bisectors of and meet DC at P, prove that i. =90^ , ii. AD = DP and PB = PC = BC, iii. DC = 2AD.

ABCD is a parallelogram in which DA = 60^ and bisectore of A and B meets DC at P. i. In a parallelogram, opposite angles are equal. So, = =60^ In a parallelogram the sum of all the four angles is 360^ + + + =360^ Now, + =360^ -( + ) =360^ -(60^ +60^ )=240^ 2 =240^ [ = ] So, = = 240^ 2 =120^ Since AB | DP and AP is a transversal SO, = = 60^ 2 =30^ ...(1)[ alternate angles] Also, AB | PC and BP is a transversal. So, = But, = 2 = 120^ 2 =60^ =60^ ...(2) Now, + + =180^ [As DPC is a straight line] 30^ + +60^ =180^ =180^ -30^ -60^ =90^ ii. Since =30^ [from (1)] and = 60^ 2 =30^ So, = Now in , = ...(3) = AD [isosceles triangle, sides are equal] As =60^ [from (2)] and =60^ So, =180^ -60^ -60^ =60^ Since all angles in the are equal, it is an equilateral triangle. = PC = BC...(4) iii. = , [from (3)] = AD [isoscele striangle, sides are equal] =BC [opposite sides are equal] =PC [from (4)] =1 2 DC [ = PC is the midpoint of DC] = 2AD.

In the adjoining figure, ABCD is a parallelogram in which =60^ . …

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Question

In the adjoining figure, $\text{ABCD}$ is a parallelogram in which $\angle\text{A}=60^{\circ}.$ If the bisectors of $\angle\text{A}$ and $\angle\text{B}$ meet $DC$ at $P$, prove that
$i. \angle\text{APB}=90^{\circ},$
$ii. AD = DP$ and $PB = PC = BC$,
$iii. DC = 2AD$.

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Answer

$\text{ABCD}$ is a parallelogram in which $DA = 60^\circ$ and bisectore of $A$ and $B$ meets $DC$ at $P$.
$i.$ In a parallelogram, opposite angles are equal.
So, $\angle\text{C}=\angle\text{A}=60^{\circ}$
In a parallelogram the sum of all the four angles is $360^\circ $
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^{\circ}$
Now, $\angle\text{B}+\angle\text{D}=360^{\circ}-(\angle\text{A}+\angle\text{C})$
$=360^{\circ}-(60^{\circ}+60^{\circ})=240^{\circ}$
$\therefore2\angle\text{B}=240^{\circ}$ $[\because\angle\text{B}=\angle\text{D}]$
So, $\angle\text{B}=\angle\text{D}=\frac{240^{\circ}}{2}=120^{\circ}$
Since $AB\ \|\ DP$ and $AP$ is a transversal
SO, $\angle\text{APD}=\angle\text{PAB}=\frac{60^{\circ}}{2}=30^{\circ}...(1)[\therefore$ alternate angles$]$
Also, $AB\ \|\ PC$ and $BP$ is a transversal.
So, $\angle\text{ABP}=\angle\text{CPB}$
But, $\angle\text{ABP}=\frac{\angle\text{B}}{2}=\frac{120^{\circ}}{2}=60^{\circ}$
$\therefore\angle\text{CPB}=60^{\circ}...(2)$
Now, $\angle\text{APD}+\angle\text{APB}+\angle\text{CPB}=180^{\circ} [$As $\text{DPC}$ is a straight line$]$
$30^{\circ}+\angle\text{APB}+60^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{APB}=180^{\circ}-30^{\circ}-60^{\circ}=90^{\circ}$
$ii.$ Since $\angle\text{APD}=30^{\circ} [$from $(1)]$
and $\angle\text{DAP}=\frac{60^{\circ}}{2}=30^{\circ}$
So, $\angle\text{APD}=\angle\text{DAP}$
Now in $\triangle\text{APD},$
$\angle\text{APD}=\angle\text{DAP}...(3)$
$\therefore\text{DP = AD} [$isosceles triangle, sides are equal$]$
As $\angle\text{CPB}=60^{\circ} [$from $(2)]$
and $\angle\text{C}=60^{\circ}$
So, $\angle\text{PBC}=180^{\circ}-60^{\circ}-60^{\circ}=60^{\circ}$
Since all angles in the $\triangle\text{PCB}$ are equal,
it is an equilateral triangle.
$\therefore\text{PB = PC = BC}...(4)$
$iii. \angle\text{DPA}=\angle\text{PAD}, [$from $(3)]$
$\therefore\text{DP = AD} [$isoscele striangle, sides are equal$]$
$=\text{BC} [$opposite sides are equal$]$
$=\text{PC} [$from $(4)]$
$=\frac{1}{2}\text{DC} [\because\text{DP = PC}\Rightarrow\text{P}$ is the midpoint of $DC]$
$\therefore\text{DC = 2AD.}$