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BT = 2PT if <BAT = < BCS = 10^o in 40^o-70^o-70^o triangle

Source: 2011 Saudi Arabia Pre-TST November 1.4

December 31, 2021
equal anglesgeometryequal segments

Problem Statement

Let ABCABC be a triangle with AB=ACAB=AC and ∠BAC=40o\angle BAC = 40^o. Points SS and TT lie on the sides ABAB and BCBC, such that ∠BAT=∠BCS=10o\angle BAT = \angle BCS = 10^o. Lines ATAT and CSCS meet at PP. Prove that BT=2PTBT = 2PT.
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