Question

Answer it correct please...? ​

Answer 1

$\huge\underline\mathtt\purple{Solution:}$

⇒ We have AB is a line segment and P,Q are points on opposite sides of AB such that,

AP = BP _____(1)

AQ = BQ_____(2)

We have to prove that PQ is perpendicular bisector of AB.

Now consider ΔPAQ and ΔPBQ,

We have, AP = BP [from(1)]

AQ = BQ [From(2)]

and, PQ = PQ [common side]

[By, SSS congruence rule],

⇒ ΔPAQ ≅ ΔPBQ_____(3)

Now, we can observe that ΔAPB and ΔABQ are isosceles triangles. (From 1 and 2)

⇒ ∠PAB = ∠PBA and ∠QAB =∠QBA

Now consider ΔPAC and PB,

C is the point of intersection of AB and PQ.

PA = PB [from(1)]

∠APC = ∠BPC [from(3)]

PC = PC [common side]

So, from SAS congruency of triangle

ΔPAC≅ ΔPBC

⇒ AC = CB and ∠PCA =∠PCB ____(4)

[ ∵ Corresponding parts of congruent triangles are equal]

And also, ACB is line segment

∠ACP + ∠BCP = 180°

But ∠ACP = ∠PCB

⇒∠ACP = ∠PCB = 90° _____(5)

We have AC = CB

⇒ C is the midpoint of AB

From (4) and (5)

We can conclude that PC is the perpendicular bisector of AB

Since C is a point on the line PQ, we can say that PQ is the perpendicular bisector of AB.

:)

Answer 2

Step-by-step explanation:

We have AB is a line segment and P,Q are points on opposite sides of AB such that,

AP = BP _____(1)

AQ = BQ_____(2)

We have to prove that PQ is perpendicular bisector of AB.

Now consider ΔPAQ and ΔPBQ,

We have, AP = BP [from(1)]

AQ = BQ [From(2)]

and, PQ = PQ [common side]

[By, SSS congruence rule],

⇒ ΔPAQ ≅ ΔPBQ_____(3)

Now, we can observe that ΔAPB and ΔABQ are isosceles triangles. (From 1 and 2)

⇒ ∠PAB = ∠PBA and ∠QAB =∠QBA

Now consider ΔPAC and PB,

C is the point of intersection of AB and PQ.

PA = PB [from(1)]

∠APC = ∠BPC [from(3)]

PC = PC [common side]

So, from SAS congruency of triangle

ΔPAC≅ ΔPBC

⇒ AC = CB and ∠PCA =∠PCB ____(4)

[ ∵ Corresponding parts of congruent triangles are equal]

And also, ACB is line segment

∠ACP + ∠BCP = 180°

But ∠ACP = ∠PCB

⇒∠ACP = ∠PCB = 90° _____(5)

We have AC = CB

⇒ C is the midpoint of AB

From (4) and (5)

We can conclude that PC is the perpendicular bisector of AB

Since C is a point on the line PQ, we can say that PQ is the perpendicular bisector of AB.We have AB is a line segment and P,Q are points on opposite sides of AB such that,

AP = BP _____(1)

AQ = BQ_____(2)

We have to prove that PQ is perpendicular bisector of AB.

Now consider ΔPAQ and ΔPBQ,

We have, AP = BP [from(1)]

AQ = BQ [From(2)]

and, PQ = PQ [common side]

[By, SSS congruence rule],

⇒ ΔPAQ ≅ ΔPBQ_____(3)

Now, we can observe that ΔAPB and ΔABQ are isosceles triangles. (From 1 and 2)

⇒ ∠PAB = ∠PBA and ∠QAB =∠QBA

Now consider ΔPAC and PB,

C is the point of intersection of AB and PQ.

PA = PB [from(1)]

∠APC = ∠BPC [from(3)]

PC = PC [common side]

So, from SAS congruency of triangle

ΔPAC≅ ΔPBC

⇒ AC = CB and ∠PCA =∠PCB ____(4)

[ ∵ Corresponding parts of congruent triangles are equal]

And also, ACB is line segment

∠ACP + ∠BCP = 180°

But ∠ACP = ∠PCB

⇒∠ACP = ∠PCB = 90° _____(5)

We have AC = CB

⇒ C is the midpoint of AB

From (4) and (5)

We can conclude that PC is the perpendicular bisector of AB

Since C is a point on the line PQ, we can say that PQ is the perpendicular bisector of AB.