Question

if two non parallel lines of a trapezium are equal, prove that it is cyclic

Answer 1

Hi there!

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For solutions, Refer to the attached picture.
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Let's see some related topics :

⚫ Circle : The collection of all the points, which are at a fixed distance from a fixed point in a plane, is called a circle.


⚫ Radius : A line joining the centre to the Circumference of the circle, is called radius of a circle.


⚫ Secant : A line intersecting a circle at any two points, is called secant.


⚫ Diameter : A chord passing through the point of the circle, is called diameter. It is the longest chord.


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Answer 2

Hello mate =_=

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Solution:

It is given that ABCD is a trapezium with AB∥CD and AD=BC

We need to prove that ABCD is a cyclic quadrilateral.

Construction: Draw AM⊥CD and BN⊥CD

In ∆AMD and ∆BNC, we have

AD=BC            (Given)

∠AMD=∠BNC          (Each equal to 90°)

AM=BN        (Distance between two parallel lines is constant.)

Therefore, by RHS congruence rule, we have ∆AMD≅∆BNC

⇒∠D=∠C        (Corresponding parts of congruent triangles are equal)   ........ (1)

We also have ∠A+∠D=180′      (Co-interior angles, AB∥CD)     ......... (2)

From (1) and (2), we can say that ∠A+∠C=180°

⇒ ABCD is a cyclic quadrilateral.

(If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.)

I hope, this will help you.

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