Question

ABCD is a rectangle in which diagonal AC bisects $\angle A$ as well as $\angle C$. Show that: Diagonal BD bisects $\angle B$ as well as $\angle D$.
[2 marks]

Answer 1

Let us join BD.
In $\Delta$ BCD,
BC=CD (sides of a square are equal to each other)
$\angle CDB=\angle CBD$(Angles apposite to equal sides are equal)
However, $\angle CDB=\angle ABD$ ( Alternate interior angles for AB || CD)
$\angle CBD=\angle ABD$
$\Rightarrow BDbisects\angle B.$
[1 mark]

Also, $\angle CBD=\angle ADB$ ( Alternate interior angles for BC || AD)
$\angle CDB=\angle ABD$
$\Rightarrow$ BD bisects $\angle$D
[1 mark]