Question

prove that altitude of a triangle are concurrent.

Answer 1

Given a triangle ABC having altitudes: AY, BX and CZ we want to show that they meet in one point.
Each of vertices of triangle constructs a line parallel to opposite side of triangle forming triangle HIJ.
JA = BC since JACB is parallelogram,
AI = BC since ABCI is parallelogram,
Hence JA = AI
Now AY is perpendicular to JI since AY is perpendicular to BC and BC is parallel to JI
So AY is the perpendicular bisector of JI.
Similarly BX is perpendicular bisector of JH and CZ the perpendicular bisector of IH.

Therefore AY, BX and CZ are concurrent as we know that the perpendicular bisectors of the sides of a triangle are concurrent at a point called the circumenter C.

Answer 2

Hlo mate here is ur ans...

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