Name: Theorems Related to Chords and Subtended Angle
Uploaded: 2019-07-19
Duration: 32 min 30 s
Description: In this video, we will study the theorem on equal chords that are equidistant from the origin and proof, examples and constructions etc, in detail.

Question

Two diameters of a circle intersect each other at right angles. Prove that the quadrilateral formed by joining their end points is a square.

Accepted Answer

Given data: Two diameters of a circle intersect each other at right angles.

Construction: Consider the figure given below.

Let $A C$ and $B D$ are two diameters of the circle, which intersect at at right angle.

As diameter intersects only at centre of circle then $O$ is centre of the circle.

So, $O A = O B = O C = O D [Radius]$

This means that $△ A O B, △ A O D, △ B O C$ and $△ C O D$ are isosceles triangles.

For $Δ D O C$ ,

$\begin{array}{l} ∠ O = 90 ° . .. (1) \\ ∠ O D C = ∠ O C D . .. (2) \end{array}$

Also

$\begin{matrix} ∠ O D C + ∠ O C D + ∠ D O C = 180 ° \\ ∠ O D C + ∠ O D C + 90 ° = 180 ° [Eq. (1) & (2)] \\ 2 ∠ O D C = 90 ° \\ ∠ O D C = 45 ° \end{matrix}$

Similarly for $Δ B O C$ , $∠ O C B = ∠ O B C = 45 °$

Thus,

$\begin{matrix} ∠ C = ∠ O C D + ∠ O C B \\ = 45 ° + 45 ° \\ = 90 ° \end{matrix}$

Similarly all angles in the quadrilateral are equal to $90^{o}$ .

$∠ A = ∠ B = ∠ C = ∠ D = 90 ° . .. (3)$

For $Δ A B C$ and $Δ A D C$ ,

$∠ B = ∠ D [90 °] A C = A C [∵ common side] ∠ O C D = ∠ O C B [45 °]$

Above conditions satisfy ASA rule for $△ A B C$ and $△ A D C$ , so both are congruent triangles.

So, $\begin{array}{l} A B = A D \\ B C = C D \end{array}$

Hence, $A B = A D = B C = C D . .. (4)$

From equations $(3)$ and $(4)$ , quadrilateral $A B C D$ is a square.

Thus, it is proved that the quadrilateral formed by joining their end points is a square.

Two diameters of a circle intersect each other at right angles. Prove that the quadrilateral formed by joining their end points is a square.

Important Questions on Circles

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