Question

In a triangle PQR, the internal bisectors of angles Q and R meet at A and the external bisectors of the angles Q and R meet at B. Prove that: ∠QAR + ∠QBR = 180°.

Answer 1

By exterior angle property,
∠RQS = ∠P + ∠R and ∠QRT = ∠P + ∠Q
Since QB bisects  ∠RQS,
`∠"BQR" = (1)/(2)∠"RQS" = (1)/(2)(∠"P" + ∠"R")`
Also RB bisects ∠QRT,
`∠"BRQ" = (1)/(2)∠"QRT" = (1)/(2)(∠"P" + ∠"Q")`
In ΔQBR,
∠QBR + ∠BRQ + ∠BQR = 180°
⇒ `∠"QBR" + (1)/(2)(∠"P" + ∠"Q") + (1)/(2)(∠"P" + ∠"R")` = 180°

⇒ `∠"QBR" + (1)/(2)(∠"P" + ∠"Q" + ∠"P" + ∠"R")` = 180°

⇒ `∠"QBR" + (1)/(2)(∠"P" + 180°)` = 180°   ....[∠P + ∠Q + ∠R = 180°]
⇒ 2∠QBR + ∠P + 180° = 360°
⇒ 2∠QBR = 180° - ∠P                               ....(i)

Since QB bisects ∠PQR,
∠AQR = `(1)/(2)∠"PQR"`
Also RA bisects ∠PRQ,
∠QRA = `(1)/(2)∠"PRQ"`
In ΔAQR,
∠AQR + ∠QRA + ∠QAR = 180°
⇒ `(1)/(2)∠"PQR" + (1)/(2)∠"PRQ" + ∠"QAR"` = 180°

⇒ `(1)/(2)(∠"PQR" + ∠"PRQ") + ∠"QAR"` = 180°
⇒ ∠PQR + ∠PRQ + 2∠QAR = 360°
⇒ 2∠QAR = 360° - ∠PQR - ∠PRQ
⇒ 2∠QAR = 180° + (180 - ∠PQR - ∠PRQ)
⇒ 2∠QAR = 180° + ∠P                              ....(ii)

Adding (i) and (ii)
⇒ 2∠QAR + 2∠QBR = 180° + ∠P + 180° - ∠P
⇒ 2∠QAR + 2∠QBR = 360°
⇒ ∠QAR + ∠QBR = 180°.