Question

let P be an interior point of a triangle ABC . let Q and R be the reflections of P in AB and AC, respectively.If Q,A,R are collinear then angle A equals ​

Answer 1

Answer:

let P be an interior point of a triangle ABC . let Q and R be the reflections of P in AB and AC, respectively.If Q,A,R are collinear then angle A equals ​= 90°

Step-by-step explanation:

 Let M be the intersection point of AB & PQ

in Δ AMQ & Δ AMP

∠M = 90° QM = PM  & AM common

Δ AMQ ≅ Δ AMP

=> ∠AQP =  ∠APQ

Similarly   Let N be the intersection point of AC & PR

=> ∠ARP =  ∠APR

in ΔQRP

∠RQP + ∠QRP + ∠QPR = 180°

∠RQP = ∠AQP

∠QRP = ∠ARP

∠QPR = ∠APQ + ∠APR = ∠AQP + ∠ARP

=> ∠AQP + ∠ARP + ∠AQP + ∠ARP  = 180°

=> 2(∠AQP + ∠ARP) = 180°

=> ∠AQP + ∠ARP = 90°

=>  ∠QPR = 90°

Now in quadrilateral

AMPN

∠A + ∠M + ∠P + ∠ N = 360°

∠M = ∠AMP = 90°

∠N = ∠ANP = 90°

∠P = ∠MPN = ∠QPR = 90°

=> ∠A + 90° + 90° + 90° = 360°

=>  ∠A = 90°