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ext{ extdollar}OCD ext{ extdollar} is a triangle. ext{ extdollar}A ext{ extdollar} is a point on ext{ extdollar}OC ext{ extdollar} and ext{ extdollar}B ext{ extdollar} is a point on ext{ extdollar}OD ext{ extdollar}. ext{ extdollar}OA = a ext{ extdollar} and ext{ extdollar}OB = b ext{ extdollar}. ext{ extdollar}OA = OC ext{ extdollar} and ext{ extdollar}OB: BD = 33 : 22 ext{ extdollar}. ext{ extdollar}X ext{ extdollar} is a point on ext{ extdollar}OC ext{ extdollar} such that ext{ extdollar}BX ext{ extdollar} is parallel to ext{ extdollar}DC ext{ extdollar}. Find ext{ extdollar}XD ext{ extdollar}. Give your answer as simply as possible in terms of ext{ extdollar}a ext{ extdollar} and ext{ extdollar}b ext{ extdollar}

Full solution

Q. ext{ extdollar}OCD ext{ extdollar} is a triangle. ext{ extdollar}A ext{ extdollar} is a point on ext{ extdollar}OC ext{ extdollar} and ext{ extdollar}B ext{ extdollar} is a point on ext{ extdollar}OD ext{ extdollar}. ext{ extdollar}OA = a ext{ extdollar} and ext{ extdollar}OB = b ext{ extdollar}. ext{ extdollar}OA = OC ext{ extdollar} and ext{ extdollar}OB: BD = 33 : 22 ext{ extdollar}. ext{ extdollar}X ext{ extdollar} is a point on ext{ extdollar}OC ext{ extdollar} such that ext{ extdollar}BX ext{ extdollar} is parallel to ext{ extdollar}DC ext{ extdollar}. Find ext{ extdollar}XD ext{ extdollar}. Give your answer as simply as possible in terms of ext{ extdollar}a ext{ extdollar} and ext{ extdollar}b ext{ extdollar}
  1. Triangle Similarity: Since OA=OCOA = OC and triangles OADOAD and OCBOCB are similar (by AA similarity), we have ODOC=OBOA\frac{OD}{OC} = \frac{OB}{OA}.
  2. Ratio Calculation: Given OB:BD=3:2OB:BD = 3:2, we can express BDBD as 25\frac{2}{5} of OBOB, so BD=2b5BD = \frac{2b}{5}.
  3. Expression for OD: Since OD=OB+BDOD = OB + BD, we can write OD=b+2b5=7b5OD = b + \frac{2b}{5} = \frac{7b}{5}.
  4. Proportion Calculation: Using the similarity of triangles, we have OD:OC=OB:OAOD:OC = OB:OA, so (7b5):a=b:a(\frac{7b}{5}):a = b:a.
  5. Error Identification: From the proportion, we get 7b5=b\frac{7b}{5} = b, which is not possible since bb cannot be equal to 7b5\frac{7b}{5} unless bb is zero, which is not the case here. We made a mistake.

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