prove that a cyclic rhombus is a square
Answers
Answer:
The proof is explained step-wise below :
Step-by-step explanation:
A rhombus is a quadrilateral with all sides of equal length.
But a square has not only all sides equal but also the measure of all interior angles are right angles.
So, to show : any rhombus is a square, we need to show any angle of a rhombus is right angle.
In the figure,diagonal BD is angular bisector of angle B and angle D.
In ΔABC and ΔBCD,
AD = BC (sides of rhombus are equal)
AB = CD (sides of rhombus are equal)
BD = BD (common side)
△ABC ≅ △BCD. (SSS congruency postulate)
In the figure,
2a + 2b = 180° (as, opposite angles of a cyclic quadrilateral are always supplementary)
2(a + b)=180°
a+b=90°
In △ABC,
⇒ ∠A = 180°- (a + b)
= 180°-90°
= 90°
Therefore,one of the interior angle of rhombus is 90°
Hence, rhombus inscribed in a circle is a square.
Hence Proved.