Question

In Δ ABC, BD⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180° − A.

Answer 1

In the given ΔABC, BD⊥AC and CE⊥AB.

We need prove ∠BOC = 180° - ∠A

Here, in ΔBDC, using the exterior angle theorem, we get,

∠BDA = ∠DBC + ∠DBC

90 = ∠DBO + ∠DBC       ................ (1)

Similarly, in ΔEBC, we get,

∠AEC = ∠EBC + ∠ECB

90 = ∠EBC ∠ECB     ................ (2)

Adding (1) and (2), we get,

90 + 90 = ∠DBC + ∠DCB + ∠EBC + ∠ECB

180 = (∠DCB +∠EBC) + (∠DBC+∠ECB)................ (3)

Now, on using angle sum property, 

In ΔABC, we get,

∠BAC + ∠ABC + ∠ACB = 180

∠ABC + ∠ACB = 180 - ∠BAC

This can be written as,

∠EBC + ∠DCB = 180 - ∠A     ......... (4)

Similarly, using angle sum property in ΔOBC, we get,

∠BOC + ∠OBC + ∠OCB = 180

∠OBC + ∠OCB = 180 - ∠BOC

This can be written as,

∠DBC + ∠ECB = 180 - ∠BOC   ......... (5)

Now, using the values of (4) and (5) in (3), we get,

         180 = 180 - ∠BOC ........... (5)

          180 = 360 - ∠A - ∠BOC

     ∠BOC = 360 - 180 - ∠A

    ∠BOC = 180 - ∠A

Therefore, ∠BOC = 180 - ∠A.

Hence proved