This is a fairly easy question on the properties of isosceles and right angled triangles.
Triangle ABC is an isosceles triangle with AC = CB. Therefore, AB is the unequal side and angle ACB is the unequal angle.
Let D be the point where the perpendicular from C meets AB. Then, triangles ACD and triangles BCD are right angled triangles with the right angle at D.
In an isosceles triangle, the perpendicular drawn from the vertex containing the unequal angle bisects the unequal side. Additionally, it also bisects the unequal angle. In short, this line acts as the perpendicular bisector of the unequal side and as the angle bisector of the unequal angle.Therefore, in the isosceles triangle ABC, AD = DB. In right angled triangle ACD, the hypotenuse AC = 4 and the perpendicular CD = 3. Using Pythagoras theorem, \({AC}^2 = {AD}^2 + {CD}^2\). Substituting the values of AC and CD in the equation above, we can calculate AD to be √7.
Since triangle BCD is exactly the same as triangle ACD (which is to say that triangle ACD is congruent with congruent BCD), it’s not hard to figure out that DB will also be √7.
AB = AD + DB. Therefore, AD = √7+ √7 = 2√7.
The correct answer option is C.
The fact that AD and DB are equal also ties in with angle bisector theorem.
The angle bisector theorem states “The angle bisector of an interior angle of a triangle divides the opposite side in the ratio of the arms of the angle bisected by it”.In this question, CD is the angle bisector of inteior angle ACB (as per the property of an isosceles triangle). The arms of this angle ACB are AC and CB which are equal and hence in the ratio of 1:1. You can now observe that the angle bisector is bisecting the opposite side AB in the ratio of 1:1 i.e. AD = DB, which is what we proved.
Hope that helps!
_________________