54 Use theorem to write 2 H Pythagoras’ expressions for z2. Triangle ABC: = + (s + z2 r2 y)2 Triangle ABD: = + (x + z2 w2 t)2 P1 P1 for combining expressions So + (s + = + (x + r2 y)2 w2 t)2 Now = r2 x2 s2 – and = w2 y2 – t2 P1 P1 for substitution Substitute for and in the equation: r2 w2 + (s + = + (x + x2 – s2 y)2 y2 – t2 t)2 P1 P1 for multiplying out to get individual terms Multiply out the brackets: + + 2sy x2 s2 s2 + y2 – P1 P1 for simplifying + + 2xt + = y2 – t2 x2 t2 Simplify: + 2sy + = + + 2xt x2 y2 y2 x2 2sy = 2xt = sy xt P1 P1 for clear explanation Alternatively: Assume T is the intersection of AD and BC. ACT is similar to BDT as both contain the same angles. (Angles ATC C1 for = and BTD are vertically opposite and so C1 C1 for clear communication of proof are equal.) Hence by similarity = C1 So = 7 xt sy
54 Use Pythagoras’ theorem to write expressions for z2. Triangle ABC: z2 = r2 + (s + y)2 Triangle ABD: z2 = w2 + (x + t)2 So r2 + (s + y)2 = w2 + (x + t)2 Now r2 = x2 – s2 and w2 = y2 – t2 Substitute for r2 and w2 in the equation: x2 – s2 + (s + y)2 = y2 – t2 + (x + t)2 Multiply out the brackets: x2 – s2 + s2 + 2sy + y2 = y2 – t2 + x2 + 2xt + t2 Simplify: x2 + 2sy + y2 = y2 + x2 + 2xt 2sy = 2xt sy = xt Alternatively: Assume T is the intersection of AD and BC. ACT is similar to BDT as both contain the same angles. (Angles ATC and BTD are vertically opposite and so are equal.) Hence by similarity = So xt = sy P1 P1 P1 P1 P1 C1 C1 2 P1 for combining expressions P1 for substitution P1 for multiplying out to get individual terms P1 for simplifying P1 for clear explanation C1 for = C1 for clear communication of proof H 7
Higher_Skills_Geometry_MarkScheme_Edexcel.docx