Math, asked by kalyanmaddipati45, 9 months ago

in the following figure angleB =90 BD perpendicular AC prove that BC^2/AB^2=DC/AD​

Answers

Answered by dheerajk1912
0

To Prove:- \mathbf{\frac{BC^{2}}{AB^{2}}=\frac{DC}{AD}}

Step-by-step explanation:-

  • Here ΔABC is a right angle triangle.
  • We know that on drawing a altitude from right angle vertices to hypotenuse , it divide into two right angle triangle. Each right angle triangle is similar to original right angle triangle.
  • From ΔACB and ΔABD , which are similar to each other.

        Means

        \mathbf{ \Delta  ACB\sim \Delta ABD}

        So ratio of corresponding side are also equal.

        \mathbf{\frac{AB}{AC}=\frac{AD}{AB}}

        ⇒\mathbf{AB^{2}=AC\times AD}        ...1)

  • From ΔCAB and Δ CBD , which are similar to each other.

        Means

        \mathbf{ \Delta  CAB\sim \Delta CBD}

       So ratio of corresponding side are also equal.

       \mathbf{\frac{BC}{AC}=\frac{DC}{BC}}

      ⇒\mathbf{BC^{2}=AC\times DC}        ...2)

  • From equation 2) ÷ equation 1)

       \mathbf{\frac{BC^{2}}{AB^{2}}=\frac{AC\times DC}{AC\times AD}}

       So

       \mathbf{\frac{BC^{2}}{AB^{2}}=\frac{DC}{AD}}    Proved

Similar questions