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AMBQID 64 ABC Right Isosceles Triangle BD = BD Find Theta


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


A. 22.5 Degree

A. 30 Degree

A. 7.5 Degree

D. 15 Degree



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 = a\sqrt{2} by using Concept 2.

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

Given that BD = BC. Thus BD = a\sqrt{2}.

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 = \frac{a}{\sqrt{2}}.

In right triangle ABM, AM = \frac{a}{\sqrt{2}} using concept 2.

Since the lines AD and MN are parallel, then AM = DN by using concept 1. Thus, DN = \frac{a}{\sqrt{2}}. 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.


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