Question

show that the diagonals of a rhombus divide it into four congruent triangles

Answer 1

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hope the attachment will help un

Answer 2

Step-by-step explanation:

Correct Question:

Show that the Diagonals of a Rhombus Divide it into 4 Equal congruent Triangles.

AnswEr:

Let us Consider that ABCD is a Rhombus.

We know that in Rhombus all sides are equal.

So, AB = BC = CD = AD ____eq(1)

Diagonal AC divides the Rhombus in two Triangles.

Now, In ∆ ABC & ∆ CDA

$\longrightarrow\sf\; AB = CD (Equal\; sides)$

$\longrightarrow\sf\; BC = DA (Equal \;sides)$

$\longrightarrow\sf\; AC = AC (Common)$

By Side Side Side (SSS) Criteria

∆ ABC ~ ∆CDA

$\rule{150}3$

In Isosceles triangle two sides are equal.

In Isosceles triangle two sides are equal. ∆ABC & ∆CDA have two equal sides so they are Isosceles triangle.

Since, ∠DAC = ∠DCA = ∠BCA = ∠BAC = $\sf\dfrac{DAB}{2}$

As BD is Diagonal Similarly we can prove that

∆ ABD ~ ∆ CBD

And, ∠ADB = ∠CDB = ∠ABD = ∠CBD = $\sf\dfrac{ABC}{2}$

Here, we can see that E is the point where Diagonals AC & BD intersect each other.

Since, ∠DAE = ∠BAE = ∠BCE = ∠DCE = $\sf\dfrac{ABC}{2}$

And, ∠ABC = ∠CBE = ∠CDE = ADE = $\sf\dfrac{ABD}{2}$

AB = BC = CD = AD (From EQ.1)

The Four triangles Are Congruent by Angle Angle Side Criteria.

The Diagonals of a Rhombus Divide it into 4 Equal congruent Triangles.

Hence Proved!

$\rule{150}3$ohoh