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ABCD is a rectangle in which diagonal BD bisects \angle B. Show that ABCD is a square. 

Answers (1)

Solution.
Given: In a rectangle ABCD, diagonal BD bisects B
To Prove: ABCD is a square
Construction: Join AC

Proof:
Given that ABCD is a rectangle. So all angles are equal to 90^{\circ}
Now, BD bisects \angle B
\angle DBA=\angle CBD

Also,
\angle DBA+\angle CBD=90^{\circ}  

So,
2\angle DBA=90^{\circ}


\angle DBA=45^{\circ}
In \triangle ABD,
\angle ABD+\angle BDA+\angle DAB=180^{\circ}  
(Angle sum property)
45^{\circ}+\angle BDA+90^{\circ}=180^{\circ}
\angle BDA=45^{\circ}
In \triangle ABD,
AD = AB (sides opposite to equal angles in a triangle are equal)
Similarly, we can prove that BC = CD
So, AB = BC = CD = DA
So ABCD is a square.
Hence proved

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