# <span lang="EN-US

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).

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I