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The proof is fairly long but relatively straightforward. You may find it easier to follow if you have a diagram: unfortunately, the support for graphics on this browser are hopelessly inadequate.

Suppose you have a rhombus ABCD so that AB = BC = CD = DA. Also AB DC and AD BC.

Suppose the diagonals of the rhombus meet at P.

Now AB DC and BD is an intercept. Then angle ABD = angle BDC.

Also, in triangle ABD, AB = AD. therefore angle ABD = angle ADC.

while in triangle BCD, BC = CD so that angle DBC = angle BDC.

Similarly, it can be shown that angle BAC = angle CAD = angle DCA = angle ACB.

Now consider triangles ABP and CBP. angle ABP (ABD) = angle CBP ( CBD or DBC),

sides AB = BC

and angle BAP (BAC) = angle BCP (BCA = ACB).

Therefore, by SAS, the two triangles are congruent.

In the same way, triangles BCP and CPD can be shown to congruent as can triangles CPD and DPA. That is, all four triangles are congruent.

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It is proven by the fact that a rhombus has 4 equal sides and that its diagonals intersect each other at right angles thus producing 4 congruent right angle triangles.

Q: How do you prove that the diagonals in a rhombus divide the rhombus into four congruent triangles?

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Multiply the diagonals and divide by 2

There are 10 possible diagonals drawn from one vertex of the 13-gon which divide it into 11 nonoverlapping triangles.

We know that diagonals of parallelogram bisect each other. Therefore, O is the mid-point of AC and BD. BO is the median in ΔABC. Therefore, it will divide it into two triangles of equal areas. Area (ΔAOB) = Area (ΔBOC) ... (1) In ΔBCD, CO is the median. Area (ΔBOC) = Area (ΔCOD) ... (2) Similarly, Area (ΔCOD) = Area (ΔAOD) ... (3) From equations (1), (2), and (3), we obtain Area (ΔAOB) = Area (ΔBOC) = Area (ΔCOD) = Area (ΔAOD) Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of equal area.

It is the line joining the midpoints of two sides of a polygon - usually a triangle. This line will be parallel to the third side. The three median-median lines will divide any triangle into 4 congruent triangles that are similar to the original.It is the line joining the midpoints of two sides of a polygon - usually a triangle. This line will be parallel to the third side. The three median-median lines will divide any triangle into 4 congruent triangles that are similar to the original.It is the line joining the midpoints of two sides of a polygon - usually a triangle. This line will be parallel to the third side. The three median-median lines will divide any triangle into 4 congruent triangles that are similar to the original.It is the line joining the midpoints of two sides of a polygon - usually a triangle. This line will be parallel to the third side. The three median-median lines will divide any triangle into 4 congruent triangles that are similar to the original.

When you bisect something you divide it into two congruent (equal) pieces

Related questions

It is a rhombus because its diagonals meet at right angles.

Yes

Yes.

Yes, they do.

Yes, draw the two diagonals. This will divide the rhombus into 4 identical triangles.

0.5

No, the diagonals of a parallelogram do not necessarily bisect the angles. The diagonals of a parallelogram divide it into four congruent triangles, but they do not necessarily bisect the angles of those triangles.

Only the square has.

If you multiply the lengths of the two diagonals, and divide by 2, you get the area of a rhombus. How does this work: Call the diagonals A & B for clarity. Diagonal A will split the rhombus into 2 congruent triangles. Looking at one of these triangles, its base is the diagonal A, and its height is 1/2 of diagonal B. So the area of one of the triangles is (1/2)*base*height = (1/2)*A*(B/2) = A*B/4. The other triangle has the same area, so the two areas together make up the whole rhombus = 2*(A*B/4) = A*B/2.

A square.

Multiply the diagonals and divide by 2

A trapezoid can be divided into 2 triangles but they are not normally congruent to each other.