畫線三角形問題

令 T 為 PR 右方一點使得 △PTR 為等邊△ ,
圖片參考：http://imgcld.yimg.com/8/n/HA04628698/o/701204110125713873392600.jpg

則 ∠SPT
= ∠SPR + ∠RPT = (180° - 80°*2) + 60° = 80° = ∠PQR
PS = QR (已知)
PQ = PR (已知) = PT (等邊△之兩邊)∴ △PQR ≌ △TPS (S.A.S)
故對應邊 TS = PR = TR(等邊△之兩邊)
∴ △TSR 為等腰△ ,
∠TRS
= (180° - ∠STR) / 2
= ( 180° - (∠PTR - ∠PTS) ) / 2
= ( 180° - (60° - 20°) ) / 2
= 70°
於是
∠PRS = ∠TRS - ∠TRP = 70° - 60° = 10°