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Math Help - right angled triangle

  1. #1
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    right angled triangle

    The diagram shows a right angked triangle where angle BAC=90 . A perpendicular is drawn from A to meet the hypothenus BC at D . Prove that triangles ABC , DBA , and DAC are similar .

    Is angle BAD = angle DAC ??
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  2. #2
    MHF Contributor red_dog's Avatar
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    No.

    In the triangle ADB: \widehat{ABD}+\widehat{DAB}=90^{\circ}

    In the triangle ABC: \widehat{ABD}+\widehat{ACD}=90^{\circ}

    Then \widehat{DAB}=\widehat{ACD}

    In the same way, \widehat{ABD}=\widehat{DAC}
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  3. #3
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    Hello, thereddevils!

    The diagram shows a right-angled triangle where angle BAC=90.
    A perpendicular is drawn from A to meet the hypothenus BC at D.
    Prove that triangles ABC, ADB, and ADC are similar .
    Code:
                    A
                    *
                   *|4 *
                  *3|     *
                 *  |        *
                *   |           *
               * 1  |            2 *
            B *  *  *  *  *  *  *  *  * C
                    D
    Is angle BAD = angle DAC? . no

    In right triangle ABC,\;\angle 1\text{ and }\angle 2 are complementary: . \angle 1 + \angle 2 \:=\:90^o

    In right triangle ADB,\;\angle 1\text{ and }\angle 3 are complementary: . \angle 1 + \angle 3 \:=\:90^o

    . . Hence: . \angle 2 = \angle 3



    So \Delta ABC and \Delta ADB have equal angles:

    . . \begin{array}{ccc}<br />
\Delta ABC & & \Delta ABD \\ \hline<br />
\angle 1 &=& \angle 1 \\<br />
\angle 2 &=& \angle 3 \\<br />
90^o &=& 90^o \end{array}


    Therefore: . \boxed{\Delta ABC \:\sim\:\Delta ABD}



    Similarly, prove that \Delta ABC \sim \Delta ADC

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