In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4

Hi ssriva2,

You can also understand this property in the following manner. Refer the diagram below:



In the diagram, ABC is an isosceles triangle with sides AB = AC. As angles opposite the equal sides are equal we have ∠B = ∠C = \(x\). Also AD is the angle bisector of ∠A, therefore ∠BAD = ∠CAD = \(\frac{y}{2}\).

We can also see from the above figure that ∠BDA = ∠CDA = \(z\) (as ∠BDA = ∠CDA = \(180 - x - \frac{y}{2}\))

We also know that in triangle ABC, \(2x + y = 180\). Similarly in triangle ABD we can write \(z + x + \frac{y}{2} = 180\) i.e. \(2z + 2x + y = 360\).

Since \(2x + y = 180\), we have \(2z + 180 = 360\) i.e. \(z = 90\).

However please note that the only the angle bisector of the non-equal angle of an isosceles triangle will be perpendicular to the opposite base i.e. the non-equal side. Also AD bisects base BC i.e. BD = DC. These properties will not apply to the equal angles and the equal sides.

In case of an equilateral triangle, we know that \(x = 60\) and \(\frac{y}{2} = 30\), hence \(z = 90\).
In an equilateral triangle since all angles and sides are equal, these properties would apply to any of the angles and sides.

Hope it's clear

Regards
Harsh