Let p be an interior point of triangle abc
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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
Asked by admin @ in Math viewed by 378 People
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
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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°
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