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Thread: Similar Triangle Definition and Problem

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    Newbie B9766's Avatar
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    Similar Triangle Definition and Problem

    Similar Triangle Definition and Problem-587369fd.jpg
    So here's an interesting problem where I have conflict with the answer and would appreciate help understanding the proof.

    In \triangle{ABC}, the point P is \frac{1}{3} down side \overline{BA}, and the point Q is \frac{1}{3} down side \overline{BC}

    Prove that \overline{PQ} is parallel to \overline{AC}

    I cannot find any definitions nor theorems that stipulate that two triangles are similar if all three corresponding sides have the same ratio - in this case: k = 3

    The book says " \triangle{PBQ}\,\sim \,\triangle{ABC} \, because of SAS ( \overline{AB}\,=\,3(\overline{PB}) \; and  \; \overline{BC} \, = \, 3(\overline{BQ})

    The SAS Theorem states: "If two sides and the included angle of one triangle are CONGRUENT to two sides and the included angle of another triangle, then these two triangles are congruent.

    It doesn't say that if two sides are SIMILAR and the included angle is congruent that the two triangles will be SIMILAR.

    Likewise, the definition of Similar Polygons states, "Two polygons are similar if, in moving one direction about one polygon, it is possible to move in some direction about the second polygon so that, 1) all side lengths encountered in turn match in the same ratio k; AND 2) all angles encountered in turn match exactly. The common ratio k of the side lengths is called the scale factor.

    It doesn't say "OR" because it would be possible to have a two, four-sided polygons with all corresponding sides in fixed ratio and yet have different angles between those two polygons.

    And the definition of an isosceles triangle states, "An isosceles triangle is a triangle with at least two sides Congruent."

    Now, it seems obvious that two triangles with three corresponding similar sides having a ratio of k (other than 1 or 0) are similar triangles and have congruent corresponding angles. But I have yet to find that definition or theorem anywhere.

    What am I missing?

    Once it's proved that the two triangles are similar it's relatively trivial to prove the two bases are parallel because it can be shown by the Similar Polygon Definition that the angles are congruent.
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    Re: Similar Triangle Definition and Problem

    Quote Originally Posted by B9766 View Post
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    So here's an interesting problem where I have conflict with the answer and would appreciate help understanding the proof.
    In \triangle{ABC}, the point P is \frac{1}{3} down side \overline{BA}, and the point Q is \frac{1}{3} down side \overline{BC}
    Prove that \overline{PQ} is parallel to \overline{AC}

    I cannot find any definitions nor theorems that stipulate that two triangles are similar if all three corresponding sides have the same ratio - in this case: k = 3.
    Look HERE at rule #3.
    Thanks from B9766
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    Newbie B9766's Avatar
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    Re: Similar Triangle Definition and Problem

    Sorry! I should have posted this in Geometry rather than New Users.
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    Re: Similar Triangle Definition and Problem

    Quote Originally Posted by B9766 View Post
    Sorry! I should have posted this in Geometry rather than New Users.
    VERY serious offence.
    You will be prosecuted with no mercy...
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