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Given BE and CF are the altitudes of the ∆ABC.P,Q are on BE and extension of CF respectively, such that BP=AC,CQ=AB, prove that AP is perpendicular to AQ
From AB perpendicular to CQ and BE perpendicular to ACangleABE=angleQCASinceAB=CQ and BP =CA,TriangleABP = TriangleQCA (S.A.S.) angleBAP = angleCQA,angleQAP = angleQAF + angleBAP = angleQAF + angleCQA = 180 °- 90 = 90.
From AB perpendicular to CQ and BE perpendicular to AC
angleABE=angleQCA
SinceAB=CQ and BP =CA,
TriangleABP = TriangleQCA (S.A.S.)
angleBAP = angleCQA,
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