ABC is a right triangle with angle B = 90º. A circle with BC as diameter meets by hypotenuse AC at point D. Prove that: i) AC x AD = AB^2 ii) BD^2 = AD x DC.
ABC is a right triangle with angle B = 90º. A circle with BC as diameter meets by hypotenuse AC at point D. Prove that: i) AC x AD = AB^2 ii) BD^2 = AD x DC.

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(C) - 19

  1. i) In 

        \[\vartriangle ABC\]

We know that

    \[\angle B\text{ }=\text{ }{{90}^{o}}~and\text{ }BC\]

is the diameter of the circle.

Hence,

    \[AB\]

is the tangent to the circle at

    \[B\]

Now, as

    \[AB\]

is tangent and

    \[ADC\]

is the secant we have

    \[A{{B}^{2}}~=\text{ }AD\text{ }x\text{ }AC\]

 

  1. ii) In 

        \[\vartriangle ADB\]

    \[\angle D\text{ }=\text{ }{{90}^{o}}\]

So,

    \[\angle A\text{ }+\angle ABD\text{ }=\text{ }{{90}^{o}}~\ldots \ldots \text{ }\left( i \right)\]

But in

    \[\vartriangle ABC,\angle B\text{ }=\text{ }{{90}^{o}}\]

    \[\angle A\text{ }+\angle C\text{ }=\text{ }{{90}^{o}}~\ldots \ldots .\text{ }\left( ii \right)\]

From (i) and (ii),

    \[\angle C\text{ }=\angle ABD\]

Now in

    \[\vartriangle ABD\text{ }and\text{ }\vartriangle CBD\]

, we have

    \[\angle BDA\text{ }=\angle BDC\text{ }=\text{ }{{90}^{o}}\]

    \[\angle ABD\text{ }=\angle BCD\]

Hence,

    \[\vartriangle ABD\text{ }\sim\text{ }\vartriangle CBD\text{ }by\text{ }AA\]

postulate

So, we have

    \[BD/DC\text{ }=\text{ }AD/BD\]

Therefore,

    \[B{{D}^{2}}~=\text{ }AD\text{ }x\text{ }DC\]