In the figure, ABCD is a parallelogram in which ∠A=60o. If the bisectors ∠A and ∠B meet DC at P, prove that AD=DP and PB=PC=BC

Question

Filo · Accepted Answer

We know that opposite angles are equal in a parallelogram

So we get

∠ C = ∠ A = 60^{o}

We know the sum of all the angles in a parallelogram is

360^{o}

It can be written as

∠ A + ∠ B + ∠ C + ∠ D = 360^{o}

So we get

∠ B + ∠ D = 360^{o} - ∠ A - ∠ C

By substituting values in the above equation we get

∠ B + ∠ D = 360^{o} - (60^{o} + 60^{o})

∠ B + ∠ D = 240^{o}

We know that

∠ B = ∠ B

So the above equation becomes

∠ B + ∠ B = 240^{o}

2∠ B = 240^{o}

By division

∠ B = ∠ D = 120^{o}

We know that

A B ∥ D P

and

A P

is a transversal

from the figure we know that

∠ A P D

and

∠ P A D

are alternate angels

∠ A P D = ∠ P A D = 60^{o} /2

∠ A P D = ∠ P A D = 30^{o} .... (1)

from (1) we know that

∠ A P D = 30^{o}

We know that

∠ D A P = 60^{o} /2 = 30^{o}

So we get

∠ A P D = ∠ D A P ....... (3)

from the figure we know that the sides of an isosceles triangle are equal

so we get

D P = A D

we know that

∠ CPB = 60^{o}

and

∠ C = 60^{o}

by sum property of triangle

we get

∠ C + ∠ CPB + ∠ PBC = 180^{o}

by substituting the values in the above equation

60^{o} + 60^{o} + ∠ PBC = 180^{o}

we get

∠ PBC = 180^{o} - 120^{o}

∠ PBC = 60^{o}

we know that all the sides of an equilateral triangle are equal

so we get

PB = PC = BC ...... (4)

therefore it is proved that

A D = D P

and

PB = PC = BC

Question Text	In the figure, $A BC D$ is a parallelogram in which $∠ A = 60^{o}$ . If the bisectors $∠ A$ and $∠ B$ meet $D C$ at $P$ , prove that $A D = D P$ and $PB = PC = BC$
Answer Type	Text solution:1
Upvotes	150

In the figure, $A BC D$ is a parallelogram in which $∠ A = 60^{o}$ . If the bisectors $∠ A$ and $∠ B$ meet $D C$ at $P$ , prove that $A D = D P$ and $PB = PC = BC$