In △PQR; PQ = PR. A is a point in PQ and B is a point in PR, so that QR = RA = AB = BP.

(i) Show that: ∠P : ∠R = 1 : 3

(ii) Find the value of ∠Q.

Δ PQR such that PQ = PR. A and B are the two points on PQ and PR, so that QR = RA = AB = BP.

(i) In Δ PQR,

PQ = PR

If two sides of a triangle are equal, then two opposite angles are always equal.

So, ∠Q = ∠R

Similarly, in Δ PAB,

AB = PB

So, ∠BAP = ∠BPA

⇒ ∠BAP = ∠P

Using exterior angle property,

⇒ ∠ABR = ∠P + ∠BAP

= ∠P + ∠P

= 2∠P

In Δ ABR,

AB = AR

So, ∠ABR = ∠ARB = 2 ∠P

Sum of all angles in a triangle ABR is 180°.

⇒ ∠ABR + ∠ARB + ∠RAB = 180°

⇒ 2∠P + 2∠P + ∠RAB = 180°

⇒ 4∠P + ∠RAB = 180°

⇒ ∠RAB = 180° - 4∠P

Sum of angles on a straight line = 180°

⇒ ∠PAB + ∠BAR + ∠RAQ = 180°

⇒ ∠P + 180° - 4∠P + ∠RAQ = 180°

⇒ - 3∠P + ∠RAQ = 0

⇒ ∠RAQ = 3∠P

In Δ ARQ,

AR = QR

So, ∠RQA = ∠QAR = 3∠P

And, we already proved, ∠Q = ∠R

So, ∠R = 3∠P

⇒ $\dfrac{∠P}{∠R} = \dfrac{1}{3}$

Hence, ∠P : ∠R = 1 : 3.

(ii) From (i), ∠P : ∠R = 1 : 3

Let ∠P be x and ∠R be 3x.

As we know ∠R = ∠Q = 3x.

In Δ PQR, sum of all angles is 180°.

∠P + ∠R + ∠Q = 180°

⇒ x + 3x + 3x = 180°

⇒ 7x = 180°

⇒ x = $\dfrac{180°}{7}$

⇒ x = 25 $\dfrac{5°}{7}$

So, ∠Q = 3x = 3 x $\dfrac{180°}{7}$

= $\dfrac{540°}{7}$

= 77 $\dfrac{1°}{7}$

Hence, the value of ∠Q = 77 $\dfrac{1°}{7}$.