CBSE Class 10 Answered
Prove that the altitudes of a triangle are concurrent
Asked by rushabhjain.avv | 01 Feb, 2019, 10:00: PM
Expert Answer
Given triangle ABC with altitudes: AE, BD and CF
Through each of the vertices of the triangle construct a line parallel to the opposite side of the triangle forming triangle PQR.
Now RA = BC since RACB is a parallelogram,
Also AQ = BC since ABCQ is a parallelogram,
Hence RA = AQ
AE is perpendicular to RQ since AE is perpendicular to BC and BC is parallel to RQ
Hence AE is the perpendicular bisector of RQ.
Similarly, BD is the perpendicular bisector of RP, and CF the perpendicular bisector of QP.
we know that the perpendicular bisectors of the sides of a triangle are concurrent at a point called the circumcenter
Through each of the vertices of the triangle construct a line parallel to the opposite side of the triangle forming triangle PQR.
Now RA = BC since RACB is a parallelogram,
Also AQ = BC since ABCQ is a parallelogram,
Hence RA = AQ
AE is perpendicular to RQ since AE is perpendicular to BC and BC is parallel to RQ
Hence AE is the perpendicular bisector of RQ.
Similarly, BD is the perpendicular bisector of RP, and CF the perpendicular bisector of QP.
we know that the perpendicular bisectors of the sides of a triangle are concurrent at a point called the circumcenter
AE, BD and CF are therefore concurrent
Answered by Arun | 07 Feb, 2019, 09:50: AM
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