Question

In the figure 6.21, $CD$ is a diameter of the circle with centre $O$. Diameter $CD$ is perpendicular to chord $AB$ at point $E$. Show that $△ABC$ is an isosceles triangle.

Answer 1

REF.Image

CD is a diameter of the circle with the center 'O'

Diameter CD is ${\perp }^{lar}$ to chord AB at point E

Also given,

to show that $\Delta ABC$ is an isosceles triangle.

Diameter = CD ($\therefore$ given)

center = O $(\therefore$ given)

CD is ${\perp }^{lar}$ to $\overrightarrow{AB}$ at E ($\therefore$ given)

so, we get from the given data,

CD $\perp$ AB $(\because$ given)

$\angle AEC=\angle BEC={90}^{\circ }$ $(\because$ from fig. (1))

CE=CE ($\because$ common)

AE=EB (common base)

as perpendicular line from the center

'O; to chord AB bisects the chord

we get,

$\Delta AEC\cong \Delta BEC$ $(\because$ as bisects)

$AC=BC(\because$ as ${\perp }^{lar}$)

as we get got side AC is equal to side BC. Now, we can say that $\Delta ABC$ is an isosceles triangle.

As in an isosceles triangle any two sides are equal.

$\therefore \Delta ABC$ is an isosceles triangle.

Hence proved.