Question

1. In the given figure, angle A = 90° and AD perpendicular on BC. Prove
that angle BAD = angle ACB.​

Answer 1

Answer:

In this figure, AD bisects ∠A

Therefore,

∠DAC = ∠DAB                 ...(Angles)

AD = DA                           ...(Common side)

∠CDA = ∠BDA                 ...(Right angle)

Hence, ΔCAD ≅ΔABD  

∴ ∠BAD = ∠ACB (corresponding parts of congruent triangles)

Step-by-step explanation:

If you like my answer please mark me as the brainliest

Answer 2

Proof of ∠BAD = ∠ACB

Given: ∠A = 90°

AD ⊥ BC

To Prove: ∠BAD = ∠ACB

Solution: ∠BAD + ∠CAD = 90° (Equation 1)

∠ACD + ∠CAD + ∠ADC = 180° (Angles of a triangle)

Since, ∠ADC = 90°

∠ACD + ∠CAD = 180° - 90°

∠ACD + ∠CAD = 90° (Equation 2)

From equations 1 and 2,

∠BAD + ∠CAD = ∠ACD + ∠CAD

∴ ∠BAD = ∠ACD

Since ∠ACD is the same as ∠ACB

Therefore, ∠BAD = ∠ACB

Hence, proved.

#SPJ3