The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that: ∠ACD + ∠BAC = 90degree - Noon Academy

The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that: ∠ACD + ∠BAC = 90degree

Class 10 | Exercise 18C | ICSE | Maths | Selina | Tangents and Intersecting Chords

The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that: ∠ACD + ∠BAC = 90degree

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(C) - 17

Solution:

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

Join

$OC$

$BCD$

is the tangent and

$OC$

is the radius.

As,

$OC\bot BD$

$\angle OCD\text{ }=\text{ }{{90}^{o}}$

$\angle OCD\text{ }+\angle ACD\text{ }=\text{ }{{90}^{o}}~\ldots .\text{ }\left( i \right)$

But, in

$\vartriangle OCA$

$OA\text{ }=\text{ }OC$

[Radii of the same circle]

Thus,

$\angle OCA\text{ }=\angle OAC$

Substituting in (i), we get

$\angle OAC\text{ }+\angle ACD\text{ }=\text{ }{{90}^{o}}$

Hence,

$\angle BAC\text{ }+\angle ACD\text{ }=\text{ }{{90}^{o}}$

i

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