# AMBQID 64:

In the given figure, the triangle ABC is a right isosceles triangle and BD = BC. Find the value of θ.

Options:

A. 22.5 Degree

A. 30 Degree

A. 7.5 Degree

D. 15 Degree

**Solution:**

To solve the above question, we need to use the following math concepts.

**Concept 1:** The perpendicular distance between the parallel lines is same at any points

**Concept 2:** In a right triangle, the Pythagoras theorem is used. It the square of the hypotenuse is the sum of the squares of the perpendicular and the square of the base.

**Concept 3:** In a right isosceles triangle, the two perpendicular lines are equal and each acute angle is 45 degree.

**Concept 4:** In an isosceles triangle, the perpendicular drawn from the vertex to opposite side bisects the side.

**Concept 5:** In trigonometry, sin(Angle) = Perpendicular / Hypotenuse.

Let the sides of right isosceles triangle AB = AC = a. Then, BC = by using **Concept 2**.

Since AB = AC = a, then angle ABC = 45 degree using concept 3.

Given that BD = BC. Thus BD = .

Now, in isosceles triangle ABC, drawn a perpendicular AM from the vertex A to the opposite side BC. Thus, BM = MC = BC/2 by using **concept 4**. Hence, BM = .

In right triangle ABM, AM = using **concept 2**.

Since the lines AD and MN are parallel, then AM = DN by using **concept 1**. Thus, DN = . Now, let angle DBN = x.

Using **concept 5**, we get sin x = DN / BD = 1/2. Thus, x = 30 degree.

Therefore, θ = 45 degree – 30 degree = 15 degree. The **correct option is D 15 degree**.