Learn UNIT IV: Geometry with Free Lessons & Tips

To construct a triangle

To construct the tangents from point

To prove that  the above equation

Let's denote the area of triangle

Given: △ABC is an isosceles triangle with AB=AC, and D is any point on BC

We'll use the Law of Cosines to prove the given relation.

Given :: solution:: for above Question

This is known as the Pythagorean Theorem, a fundamental principle in geometry. Let's denote the sides of the triangle as aa, b, and c, where c is the side opposite the right angle.

To prove that the parallelogram circumscribing a circle is a rhombus, we first need to establish some properties:

1. In a parallelogram, opposite sides are equal in length.
2. In a parallelogram, opposite angles are equal in measure.

Now, let's consider a parallelogram ABCD circumscribing a circle with center O. Let P, Q, R, and S be the points of tangency between the circle and the sides of the parallelogram, as shown below:
        P-----Q
       /       \
      /         \
     A-----------B
     |     O     |
     |           |
     D-----------C
      \         /
       \       /
        R-----S

Let's denote the midpoint of BC as M. To prove that the tangent to the circle at P bisects BC we need to show that BM=MC.

To prove that quadrilateral