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Problem 1:In the triangle ABC shown below, A'C' is
parallel to AC. Find the length y of BC' and the length x of
A'A.




Solution to Problem 1:


BA is a transversal that intersects the two parallel lines A'C' and AC,
hence the corresponding angles BA'C' and BAC are congruent. BC is also
a transversal to the two parallel lines A'C' and AC and therefore angles
BC'A' and BCA are congruent. These two triangles have two congruent
angles are therefore similar and the lengths of their sides are
proportional. Let us separate the two triangles as shown below.
We now use the proportionality of the lengths of the side to write
equations that help in solving for x and y.


(30 + x) / 30 = 22 / 14 = (y + 15) / y


An equation in x may be written as follows.


(30 + x) / 30 = 22 / 14


Solve the above for x.
420 + 14 x = 660


x = 17.1 (rounded to one decimal place).


An equation in y may be written as follows.


22 / 14 = (y + 15) / y


Solve the above for y to obtain.


y = 26.25

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Zimilar tr

  • 1. Problem 1:In the triangle ABC shown below, A'C' is parallel to AC. Find the length y of BC' and the length x of A'A. Solution to Problem 1: BA is a transversal that intersects the two parallel lines A'C' and AC, hence the corresponding angles BA'C' and BAC are congruent. BC is also a transversal to the two parallel lines A'C' and AC and therefore angles BC'A' and BCA are congruent. These two triangles have two congruent angles are therefore similar and the lengths of their sides are proportional. Let us separate the two triangles as shown below.
  • 2. We now use the proportionality of the lengths of the side to write equations that help in solving for x and y. (30 + x) / 30 = 22 / 14 = (y + 15) / y An equation in x may be written as follows. (30 + x) / 30 = 22 / 14 Solve the above for x.
  • 3. 420 + 14 x = 660 x = 17.1 (rounded to one decimal place). An equation in y may be written as follows. 22 / 14 = (y + 15) / y Solve the above for y to obtain. y = 26.25