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and AC and BC are st. lines, then ∠CEB = CDA.

75. METHOD II. If the lines or angles whose equality we wish to demonstrate are not parts of congruent triangles, we have to make them parts of congruent triangles by drawing additional lines.

Ex. 151. If the opposite sides of a quadrilateral are equal, the opposite angles are equal.

Ex. 152. If two circles whose centers are and O' intersect in A and B, then ∠A00' = ∠BOO'.

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Ex. 153. If two circles whose centers are O and O' intersect in A and B, then ∠OAO' = ZOBO'.

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76. METHOD III. If it is impossible to prove the congruence of the required pair of triangles, prove first the congruence of some other pair, or pairs, whose homologous parts will enable us to demonstrate the congruence of the original pair.

Ex. 154. If the opposite sides of a quadrilateral are equal, and a line be drawn through the midpoint of the diagonal, terminating in two sides, this line is bisected.

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Ex. 155. In the annexed diagram, if AG=GD, CG = FG, and all lines are straight lines, then BG = GE.

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Ex. 156. In the annexed diagram, if AB = AC, AE bisects LBAC, and AE is a st. line, then 2 DBE = DCE.

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D

A

Ex. 155

Ex. 157. In the annexed diagram, if AB=CD, BC = DA, and 21 = 22, and DB is a st. line, then

AE = CF.

Ex. 158. In the same diagram, if AB = CD, 23=24 and BF = DE, then 1 = 22.

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Ex. 159. Two triangles are congruent if two sides and the median to one of these sides are equal respectively to two sides and the homologous median of the other.

Ex. 160. Two isosceles triangles are congruent if an arm and the line which joins the midpoints of the arms of one triangle are equal respectively to an arm and the line which joins the midpoints of the arms of the other.

Ex. 161. If the opposite sides of a quadrilateral are equal, the diagonals bisect each other.

Ex. 162. Two triangles ABC and A'B'C' are congruent if AB = A'B', LA = LA', and angle-bisector AD = angle-bisector A'D'.

Ex. 163. If in quadrilateral ABCD, AB = BC, CD = DA, and diagonals AC and BD meet in E, then AE = EC.

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Ex. 164. If in quadrilaterals ABCD, and A A'B'C'D', AB = A'B', BC = B'C', CD = C'D', DA = D'A', and AC = A'C', then BD = B'D'.

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*Ex. 165. If on the sides of an equilateral triangle ABC, the points D, E, F are taken, so that AD = BE = CF, and E, D, and Fare joined to the opposite vertices, then A A'B'C' is equilateral.

* Ex. 166. If the opposite sides of a polygon of six sides are equal, and two of the opposite angles are equal, then all opposite angles are equal.

*Ex. 167. In the annexed diagram, if AB = AD, AC = AE, and BE and DC are st. lines, then ∠BAF = ∠ DAF.

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* See footnote on page 20.

* Ex. 168. In the annexed diagram, if ∠A = ∠B, and AF = BE, then AD = DB.

[For practical application see Ex. 3, page 285.]

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77. METHOD IV. To prove that an angle is a right angle we usually demonstrate that it is equal to its supplementary adjacent angle.

Ex. 169. The bisector of the vertex angle of an isosceles triangle is perpendicular to the base.

Ex. 170. The median to the base of an isosceles triangle is perpendicular to the base.

Ex. 171. If in quadrilateral ABCD AB = BC, and CD = DA, then BDAC.

Ex. 172. In the diagram opposite, if AC = BC, and A0 = BO, then CD AB.

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CD

AB.

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78. A point is said to be equidistant from two other points A and B if AC=BC.

C

It is not necessary that these two lines be drawn. Thus, we may say D is equidistant from A and B, even if DA and DB are not drawn. In general, two letters, as AD, mean the straight line connecting A and D.

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PROPOSITION VI. THEOREM

79. The line that joins the vertices of two isosceles triangles on the same base bisects the common base at right angles.

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Given the isos. A ABC and BCE, with the common base BC.

To prove AE is the perpendicular-bisector of BC.

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