Question

In the given figure, ABC is a triangle. BD is perpendicular to AC and CE is perpendicular to AB that meet AC and AB at D and E respectively. Prove that angle BIC= 180° - angle BAC.​

Answer 1

Given: A ∆ABC, BD⊥AC and CE⊥AB.

To Prove: ∠BIC = 180° – ∠A.

Proof:

BD⊥AC   (Given)

 

∴ ∠BDA = 90°

CE⊥AB  (Given)
∴ ∠CEA = 90°

In quadrilateral AEID,

 

∠A + ∠AEI + ∠EID + ∠IDA = 360°    (Sum of angles of quadrilateral is 360°)

 

⇒∠A + 90° + ∠EID + 90° = 360

∠A + ∠EID = 360-180°=180°
so ∠EID= 180°-∠A
but ∠EID= ∠BIC
so
∠BIC=180°-∠A

Answer 2

Answer:

Given AB=AC,BD⊥AC

From right △ADB,

By pythagorus theorem, AB

2

=AD

2

+BD

2

⟹AC

2

=AD

2

+BD

2

[∵AB=AC]

⟹(AD+DC)

2

=AD

2

+BD

2

⟹AD

2

+2.AD.DC+DC

2

=AD

2

+BD

2

⟹2.AD.DC=BD

2

−CD

2