Given In ΔABC, ∠A = 90° and AL ⊥ BC
To prove ∠BAL = ∠ACB
Proof In ΔABC and ΔLAC, ∠BAC = ∠ALC [each 90°] …(i)
and ∠ABC = ∠ABL [common angle] …(ii)
On adding Eqs. (i) and (ii), we get
∠BAC + ∠ABC = ∠ALC + ∠ABL …(iii)
Again, in ΔABC,
∠BAC + ∠ACB + ∠ABC = 180°
[sum of all angles of a triangle is 180°]
=>∠BAC+∠ABC = 1 80°-∠ACB …(iv)
In ΔABL,
∠ABL + ∠ALB + ∠BAL = 180° [sum of all angles of a triangle is 180°]
=> ∠ABL+ ∠ALC = 180° – ∠BAL [∴ ∠ALC = ∠ALB= 90°] …(v)
On substituting the value from Eqs. (iv) and (v) in Eq. (iii), we get 180° – ∠ACS = 180° – ∠SAL
=> ∠ACB = ∠BAL
Hence proved.
