In fig. ABC is right angled at A. Al is
drawn perpendicular to BC.
| Prove that LBAL = LACB
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Answer:
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.
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