Question

prove diagonals of a rhombus are perpendicular

Answer 1

Answer:

A rhombus is a parallelogram with four congruent sides.

So, all sides of rhombus ABCD are congruent.

i.e, $\overline{JM} \cong \overline{JK} \cong \overline{KL}\cong \overline{LM}$

Also, we know that the diagonals of a parallelogram bisect each other.  

Since a rhombus is a parallelogram.

By property of rhombus , if point N is the intersection of the diagonals as shown in the figure, then

$\overline{MN} \cong \overline{NK}$         .....[1]

$\overline{JN} \cong \overline{NL}$

In ΔJNM and ΔJNK

$\overline{MN} \cong \overline{NK}$     [side]    [by (1)]

$\overline{JM} \cong \overline{JK}$     [side]          [Given]

By reflexive property states that a segment is congruent to itself:

$\overline{JN} \cong \overline{JN}$   [Side]                 [Reflexive Property]

SSS(Side-Side-Side) postulates states that if three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent.

then by SSS congruence,

$\triangle JNM \cong \triangle JNK$

By CPCT [Corresponding Part of congruent triangles are congruent]

Since, JNM and JNK are corresponding angles therefore,

$\angle JNM \cong JNK$

Linear pair theorem states that two angles that form a linear pair are supplementary.

By linear pair theorem, JNM and JNK are supplementary

this mean:

$m\angle JNM +m\angle JNK =180^{\circ}$

Since, the angles are congruent i.e, $\angle JNM \cong JNK$

so;

$m\angle JNK +m\angle JNK =180^{\circ}$

or

$m\angle JNK +m\angle JNK =180^{\circ}$

2$m\angle JNK=180^{\circ}$

Simplify:

$m\angle JNK =90^{\circ}$

also; $m\angle JNM =90^{\circ}$

therefore, the diagonals of JKLM are perpendicular to each other i,e $\overline{JL} \perp \overline{MK}$