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⇒∠BAC+∠ABC = 1 80°−∠ACB ……….(iv)
In ΔABL,
∠ABL + ∠ALB + ∠BAL = 180° (Angle Sum Property of Triangle)
⇒∠ABL+ ∠ALC = 180° – ∠BAL [∴∠ALC = ∠ALB= 90°] ………(v)
On substituting the value from equations. (iv) and (v) in Eq. (iii), we get
180° – ∠ACS = 180° – ∠SAL
⇒∠ACB = ∠BAL (Proved)
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