Question

show that diagonals to a rhombus divide into four congruent triangles
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Answer 1

A rhombus has its opposite sides equal and parallel.

So, if we take the diagonal as the transversal for two opposite sides then we can get alternate interior angles.

Therefore by A.S.A congruency opposite triangles are congruent.

In a similar war to the other triangles can be done through this process.

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

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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!

$\rule{150}3$