If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.
Solution:
Consider an isosceles triangle ABC with base BC
A line ED is drawn parallel to the base BC to intersect its equal sides AB and AC.
We have to prove that the quadrilateral so formed is cyclic.
Draw a circle that passes through the points B, C, D and E.
Considering triangle ABC,
AB = AC = equal sides of an isosceles triangle
We know that the angles opposite to the equal sides are equal.
∠ACB = ∠ABC -------------------------- (1)
Given, DE || BC
We know that the corresponding angles are equal.
So, ∠ADE = ∠ACB --------------------- (2)
Adding ∠EDC on both sides in (2),
∠ADE + ∠EDC = ∠ACB + ∠EDC
We know that the linear pair of angles is always supplementary.
So, ∠ADE + ∠EDC = 180°
Now, ∠ACB + ∠EDC = 180°
This implies the sum of the opposite angles is equal to 180°
We know that the sum of the opposite angles is always supplementary in a cyclic quadrilateral.
Therefore, BCDE is a cyclic quadrilateral.
✦ Try This: In the given figure ∠AOB = 80°. The value of x is
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 11
If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.
Summary:
If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, it is proven that the quadrilateral so formed is cyclic
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