Prove that altitudes of a triangle are concurrent

#### Solution

Consider ∆ABC.

Let AP ⊥ BC and BQ ⊥ AC.

Let AP and BQ intersect at O.

Join OC and extend OC to meet AB at R.

To prove that CR is also the altitude of ∆ABC.

i.e., to prove that CR ⊥ AB

Let `bar"a", bar"b", bar"c"` be the position vectors of the points A, B, C respectively.

Consider `bar"AP" ⊥ bar"BC"`

∴ `bar"AO" ⊥ bar"BC"`

∴ `bar"AO" * bar"BC"` = 0

∴ `-bar"a"*(bar"c" - bar"b")` = 0 .......`[∵ bar"AO" = -bar"OA"]`

∴ `bar"a"*bar"c" - bar"a".bar"b"` = 0 .......(i)

Now, `bar"BQ" ⊥ bar"AC"`

∴ `bar"BO" ⊥ bar"AC"`

∴ `bar"BO" * bar"AC"` = 0

∴ `-bar"b"*(bar"c" - bar"a")` = 0 .......`[∵ bar"BO" = -bar"OB"]`

∴ `bar"b"*bar"c" - bar"b"*bar"a"` = 0 .......(ii)

Comparing equations (i) and (ii), we get

∴ `bar"a"*bar"c" - bar"a"*bar"b" = bar"b"*"c" - bar"b"*bar"a"`

∴ `bar"a"*bar"c" = bar"b"*bar"c"`

∴ `bar"a"*bar"c" - bar"b"*bar"c"` = 0

∴ `bar"c"*(bar"a" - bar"b")` = 0

∴ `-bar"c"*(bar"a" - bar"b")` = 0

∴ `bar"CO" ⊥ bar"BA"`

∴ `bar"CR" ⊥ bar"BA"`

∴ CR ⊥ BA

∴ CR is also the altitude of ∆ABC.

∴ AP, BQ, CR intersect at O.

∴ All three altitudes of ∆ABC intersect at a common point.

Thus, the altitudes of a triangle are concurrent.