In the adjoining figure, ABCD is a parallelogram in which =60^ . If the bisectors of and meet DC at P, prove that 1. =90^ , 2. AD = DP and PB = PC = BC, 3. DC = 2AD.

ABCD is a parallelogram in which DA = 60° and bisectore of A and B meets DC at P. 1. 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^ 2. 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) 3. = , [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, 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

$\angle\text{APB}=90^{\circ},$
AD = DP and PB = PC = BC,
DC = 2AD.

✓

Answer

ABCD is a parallelogram in which DA = 60° and bisectore of A and B meets DC at P.

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°

$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^{\circ}$

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 DPC is a straight line]

$30^{\circ}+\angle\text{APB}+60^{\circ}=180^{\circ}$

$\Rightarrow\angle\text{APB}=180^{\circ}-30^{\circ}-60^{\circ}=90^{\circ}$

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.

$\therefore\text{PB = PC = BC}...(4)$

$\angle\text{DPA}=\angle\text{PAD},$ [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]

$\therefore\text{DC = 2AD.}$

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