Question

in the figure ABCDE is a Pentagon with all sides of the same length and all angle of the same size. The side AB and AE extended meet the side CD extended at Pand Q. (1) Are the side of triangleBPC equal to the side of triangle EQD? why? . (2) Are the side of AP and AQ of triangle APQ equal? why?​

Answer 1

Step-by-step explanation:

Here, all angles of the polygon are equal(a regular polygon).

∴ ∠ABC = ∠AED …(eq)1

(∠PBC,∠ABC) and (∠AED,∠QED) form a linear pair.

∠PBC = 180-∠ABC (Linear angles are supplementary)

= 180-∠AED (from eq1)

∠PBC = ∠QED …(eq)2 (∠AED and ∠QED form linear pair).

∠BCD = ∠EDC …(eq)2

(∠BCD, ∠BCP) and (∠EDC,∠EDQ) form a linear pair.

∠BCP = 180-∠BCD (Linear angles are supplementary)

= 180-∠EDC (from eq2)

∠BCP = ∠EDQ …(eq)3

According to property,

If one side of a triangle and angle at its ends are equal to one side of another triangle and the angles at its ends, then the third angles are also equal and the sides opposite equal angle are equal.

In ΔPBC and Δ QED,

∠BCD = ∠EDC

∠BCP = ∠EDQ

Also, BC = DE (a regular polygon)

Hence, PC = DQ (∠PBC = ∠DEQ)

PB = EQ (∠PCB = ∠EDQ)

And ∠BPC = ∠EQD …(eq)4

Hence, the sides are equal.

ii) From eq4,

∠BPC = ∠EQD

∴ ∠APQ = ∠AQP

Hence, AP = AQ (converse of isosceles angle theorem).

I hope it will help you