Question

The diagonal $AC$ divides the rectangle $ABCD$ into two triangles. Are the two triangles congruent? If so, what are all the laws of congruency that can be used to prove it?

• A. No
• B. Yes, All congruency laws
• C. Yes, only S.A.S and S.S.S

Answer 1

The correct option is B Yes, All congruency laws


In above figure,
diagonal $AC$ divides rectangle $ABCD$ in two triangles: $△ADCand△CBA$
Now, in $△ADCand△CBA,$
$AD=CB,DC=BA$ -----(1)
[Opposite sides of rectangle are equal]
$AC=AC$ [Common side]-----(2)
From (1) & (2),

$\therefore △ADC\cong △CBA$ [By S.S.S congruency rule]

$\angle ADC=\angle CBA={90}^{\circ }$----(3)
[Each angle of rectangle is ${90}^{\circ }$]
From (1) & (3),

$\therefore △ADC\cong △CBA$ [By S.A.S congruency rule]

Similarly, All other congruency rules can be applied.
Here, $\angle DAC=\angle BCA={45}^o$
$\angle D=\angle B={90}^o$
$DC=AB$

$\therefore △ADC\cong △CBA$ by Angle-angle-side rule.

Here, $\angle DAC=\angle BCA={45}^o$
$AD=BC$
$\angle D=\angle B={90}^o$

$\therefore △ADC\cong △CBA$ by Angle-side-angle rule.

Here, $AC=AC$(common hypotenuse)
$AD=BC$(altitudes of both right triangles)
$\angle D=\angle B={90}^o$

$\therefore △ADC\cong △CBA$ by Hypotenuse-Leg congruency rule.