Question

Prove that : In a square two diagonals are equal and it bisect right angle triangle.

Answer 1

R.E.F image

Let ABCD be the square

$\Rightarrow AB=BC=CD=DA=a$

In $△ACD,$ By Pythagoras

theorem, $A{C}^2=C{D}^2+D{A}^2=2a^2$

similarly in $△BCD,B{D}^2=D{C}^2+B{C}^2=2a^2$

$\Rightarrow AC=BD$ In $△DBA$ and $△DBC,$

$DB=DB,DA=DC,BA=BC$

$\Rightarrow$ By SSS concurrency,

$△DBA\cong △DBC$

$\Rightarrow \angle DBA=\angle DBC$

$\Rightarrow$ Diagonal bisect right angle