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How do you prove the diagonals of a rhombus divide the rhombus into four congruent triangles?
Wiki User
∙ 8y ago
Take a rhombus ABCD.
A rhombus as 4 equal sides, thus AB = BC = CD = DA
Draw in 1 diagonal AC. This splits the rhombus into 2 triangles. ABC and CDA with side AB = CD, BC = DA and AC common to both triangles. Thus ABC and CDA are congruent by Side-Side-Side.
Triangles ABC and CDA are isosceles triangles since they have two equal sides (AB = BC and CD = DA) thus angles DAC = DCA = BAC = BCA. Specifically DAC = BAC. But DAC + BAC = DAB, thus DAC = BAC = ½ DAB; similarly DCA = BCA = ½ BCD = ½ DAB
Drawing in the other diagonal BD, the same arguments show triangles ABD and CDB are congruent and angles ADB = CDB = ABD = CBD = ½ ABC
Let the point where the diagonals meet be E. We now have 4 triangles ABE, BCE, CDE and DAE with equivalent angles and sides:
Angles DAE* = BAE = BCE = DCE (= ½ DAB)
Angles ABE = CBE = CDE = ADE (= ½ ABD)
Sides AB = BC = CD = DA
Thus the 4 triangles are congruent by Angle-Angle-Side.
*Angle DAE = DAC since E lines along AC; similarly for all the other angles involving point E, ie angle BCE = BCA, ADB = ADE, etc
Wiki User
∙ 8y ago
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In which parallelogram does the diagonal divide the parallelogram into two congruent right triangles?
It is a rhombus because its diagonals meet at right angles.
Are square diagonals divide it into congruent triangles?
Yes
Do the diagonals of a rhombus divide it into four triangles of equal areas?
Yes.
Do the diagonals of a rhombus divide it into four triangles of equal area?
Yes, they do.
Can you show four identical right triangles in a rhombus?
Yes, draw the two diagonals. This will divide the rhombus into 4 identical triangles.
Does 2 diagonals of a square divide it into 4 congruent triangles?
0.5
Does the diagonals of a parallelogram bisect the angles?
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.
Which type of quarilateral has diagonals that will always divide it into four congruent triangles?
Only the square has.
Where did the area of a rhombus come from?
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.
How do you prove that the diagonals in a rhombus divide the rhombus into four congruent triangles?
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 = BCand 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.
Diagonals that divide it into isosceles right triangles?
A square.
How do you calculate the area of a rhombus?
Multiply the diagonals and divide by 2
