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

Unable to understand how ab x bd = ac x dc means AD is angle bisector?

ab ac
--- = ----
bd dc

means ABD and ACD are simliar triangles , not same triangles.
Sides are in proportion, does it mean angles are same?

Kindly clarify.

It's the angle bisector theorem. Check Wiki here: https://en.wikipedia.org/wiki/Angle_bisector_theorem

Bunuel,
I have one query here.
Does the angle bisector extends 90 degree angle at base?

Sorry, I don't understand what you mean...

I hope its clear now.i have attached figure for same

Only for for an isosceles triangle (or for an equilateral), the angle bisector to base coincides with the height.

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

So based on this itself why is it E? why not D?