Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base. - Noon Academy

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Circles | Class 10 | Exercise 17C | ICSE | Maths | Selina

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Selina Solutions Concise Class 10 Maths Chapter 17 ex. 17(C) - 4

Suppose,

$\vartriangle ABC,\text{ }AB\text{ }=\text{ }AC$

and circle with

$AB$

as diameter is drawn which intersects the side

$BC\text{ }and\text{ }D$

And, join

$AD$

Proof:

It’s seen that,

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

[Angle in a semi-circle]

And,

$\angle ADC\text{ }+\angle ADB\text{ }=\text{ }{{180}^{o}}~$

[Linear pair]

Thus,

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

Now, in right

$\vartriangle ABD\text{ }and\text{ }\vartriangle ACD$

$AB\text{ }=\text{ }AC$

[Given]

$AD\text{ }=\text{ }AD$

[Common]

$\angle ADB\text{ }=\angle ADC\text{ }=\text{ }{{90}^{o}}$

Hence, by R.H.S criterion of congruence.

$\vartriangle ABD\cong \vartriangle ACD$

Now, by CPCT

$BD\text{ }=\text{ }DC$

Therefore,

$D$

is the mid-point of

$BC$

i

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