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प्रश्न
Prove that altitudes of a triangle are concurrent
उत्तर
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.
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