Name: Sum of Exterior Angles of a Polygon
Uploaded: 2019-07-19
Duration: 5 min 7 s
Description: Everyone talks about the interior angles of a polygon, but what about the external ones? In this video, you will learn about the sum of exterior angles of a...

Question

Prove that the diagonals of a rhombus bisect each other at

90 °

.

Accepted Answer

In rhombus $ABCD$ , the diagonals $AC$ and $BD$ intersect each other at $E$ .

To proof: $AE = EC, BE = DE$ and $∠ AED = 90 °$

Proof: In $△ DEC$ and $△ AEB$

$∠ ECD = ∠ EAB$ (Alternate angles)

$AB = CD$ (Opposite sides of rhombus are equal)

$∠ EDC = ∠ EBA$ (Alternate angles)

$△ DEC ≅ △ AEB (ASA)$

Corresponding parts of congruent triangles are equal.

So, $AE = EC$ and $BE = DE$ ___________ $(1)$

Therefore, the diagonals $AC$ and $BD$ bisect each other at $O$ .

Now in $△ AED$ and $△ CED$ ,

$DE = DE$ (Common side)

$90 °$ (Equal sides of rhombus)

$AE = EC$ (From $1$ )

Therefore, $△ AED ≅ △ CED$

So, $∠ AED = ∠ CED$ $(c . p . c . t)$

$∠ AED + ∠ CED = 180 °$ (Linear pair)

$\Rightarrow ∠ AED + ∠ AED = 180 °$ $[∠ AED = ∠ CED]$

$\Rightarrow 2 ∠ AED = 180 °$

$\Rightarrow ∠ AED = \frac{180 °}{2} = 90 °$

Therefore, $∠ AED = 90 °$

So, the diagonals $AC$ and $BD$ bisect each other at $90 °$ .

Hence, proved.

Prove that the diagonals of a rhombus bisect each other at $90 °$ .

Important Questions on Understanding Quadrilaterals

Learn and Prepare for any exam you want !

Prove that the diagonals of a rhombus bisect each other at 90°.

Important Questions on Understanding Quadrilaterals

Learn and Prepare for any exam you want !

Prove that the diagonals of a rhombus bisect each other at $90 °$ .