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

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

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

    In the same way, $\displaystyle \widehat{ABD}=\widehat{DAC}$
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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 $\displaystyle ABC,\;\angle 1\text{ and }\angle 2$ are complementary: .$\displaystyle \angle 1 + \angle 2 \:=\:90^o$

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

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



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

    . . $\displaystyle \begin{array}{ccc}
    \Delta ABC & & \Delta ABD \\ \hline
    \angle 1 &=& \angle 1 \\
    \angle 2 &=& \angle 3 \\
    90^o &=& 90^o \end{array}$


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



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

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