Question

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

Answer 1

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