Prove that the altitudes of a triangle are concurrent.

Hello student,

We need to prove that altitudesAD,BEandCFintersect
at one point.Let us draw the straight lineGHpassing throughthe pointCparallel to the triangle sideAB.

Similarly, we can construct lineGIpassingthrough the pointBparallel to sideAC, andthe straight lineHIpassing through the pointAparallelto the triangle sideBC. LetG,H, andIbe the intersection points of the constructed straight lines.

Now, quadrilateralABCHhas the opposite sides parallel, therefore its opposite sides are of equal length in accordance to the properties of sides of parallelograms
In particular, the sidesABandHCare of equal length:AB=HC.
Similarly, the sidesABandCGare of equal length:AB=CG(from the quadrilateralABGC). Hence,HC=CG, and the pointCis the midpoint of the segmentHG.
By the same reason, the pointAis the midpoint of the segmentHI(BC=AHfrom the quadrilateralBCHA, andBC=AIfrom the quadrilateralBCAI, which givesAH=AI).
Finally, the pointBis the midpoint of the segmentGI(AC=BGfrom the quadrilateralBGCA, andAC=BIfrom the quadrilateralIBCA, which givesBG=BI).
Thus the pointsA,BandCare the midpoints of the sides of the triangleIGH, and the altitudesAD,BEandCFof the original triangleABCare the perpendiculars to the sides of the constructed triangleIGHat the midpoints of its sides. Therefore, the segmentsAD,BEandCFintersect in one common point

Thanks and Regards

Shaik Aasif

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