Theorem 3

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Orthocentre

Definition:

The point of concurrency of the three altitudes of a triangle is called its "Orthocentre". It is generally abbreviated as 'O'.

To locate orthocentre it is sufficient to draw altitudes of any two sides of a triangle. The third altitude will then automatically pass through it.

Theorem 3

In a triangle, the three altitudes pass through the same point.

Given:

In DABC, AD, BE and CF are the altitudes.

To prove:

AD, BE and CF are concurrent (or pass through the same point)

Construction:

Through A, B and C draw lines parallel to BC, AB and AC respectively. Let these lines meet at P, Q and R forming DPQR.

Proof:

QA || BC and QC || AB by construction.

ABCQ is a parallelogram.

AQ = BC . . . (1)

Similarly BCAR is a parallelogram.

AR = BC . . . (2)

From (1) and (2),

AQ = AR . . . (3)

AD is the perpendicular bisector of RQ.

Similarly we can prove that BE is the perpendicular bisector of PR and CF is the perpendicular bisector of PQ. Thus AD, BE and CF are the perpendicular bisectors of the sides of DPQR. Hence AD, BE and CF pass through the same point. (by theorem 2)