If BE us drawn so tha angle ABE = angle DBC, complete the details of the followi
ID: 2943302 • Letter: I
Question
If BE us drawn so tha angle ABE = angle DBC, complete the details of the following proof of Ptolemy's theorem:
a) The triangles ABE and DBC are similar
AB/ BD = AE/CD
b) angle ABD = angle ABE + angle EBD = angle DBC + angle EBD = angle EBC
c)The triangles ABD and EBC are similar
AD/EC = BD/BC
d) The result of adding AB*CD=AE*BD and BC*AD=EC*BD is
AC*BD = AB*CD + BC*AD
In the Almagest, Ptolemy proved a geometrical result known today as ''Ptolemy's theorem.'' If ABCD is a (convex) quadrilateral inscribed in a circle, then the product of the diagonals is equal to the sum of the products of the two pairs of opposite sides. In symbols: AC * BD = AB * CD + BC * AD If BE us drawn so tha angle ABE = angle DBC, complete the details of the following proof of Ptolemy's theorem: a) The triangles ABE and DBC are similar AB/ BD = AE/CD b) angle ABD = angle ABE + angle EBD = angle DBC + angle EBD = angle EBC c)The triangles ABD and EBC are similar AD/EC = BD/BC d) The result of adding AB*CD=AE*BD and BC*AD=EC*BD is AC*BD = AB*CD + BC*ADExplanation / Answer
a. We shall try to find two pairs of equal angles in triangles ABE and DBC. We are given that angle ABE equals angle DBC, so we have one pair. Note that angle BAE and angle BDC are both on the circle and cut off the same arc (minor arc BC), so they must be equal as well. We therefore have two pairs of equal angles in triangles ABE and DBC, so by AA similarity, the triangles are similar. We note which angles correspond to one another - the angle opposite AE corresponds to the one opposite CD, and the one opposite AB corresponds to the one opposite BD, giving us the required equality of AB/ BD = AE/CD.
b. By the Angle Addition Postulate, angle ABD = angle ABE + angle EBD. We're given that angle ABE = angle DBC, so we can substitute, giving angle ABD = angle ABE + angle EBD = angle DBC + angle EBD. Again, we use the Angle Addition Postulate, giving the desired result of angle ABD = angle ABE + angle EBD = angle DBC + angle EBD = angle EBC.
c. In part b, we showed that angle ABD = angle DBC. angle DBC is the same as angle EBC, so we have one pair of equal angles. Angle BCA and angle BDA are on the circle and cut off the same arc (minor arc AB), so they are equal. As in part A, we use AA similarity to show that ABD and EBC are similar, and by matching the corresponding angles and using the definition of similar polygons, we get AD/EC = BD/BC.
d. Adding the two equalities that we've generated, we obtain:
AB*CD + BC*AD = AE*BD + EC*BD. Note, however, that the right hand side is equal to (AE + EC) * BD = AC*BD.
We have AC*BD = AB*CD + BC*AD, as we wanted to prove.