Question

Prove that the diagonal of a rhombus divides the rhombus into two congruent triangles. Also,
prove that the diagonal of rhombus bisects its vertex angles.
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Answer 1

Step-by-step explanation:

First, all four sides of a rhombus are congruent, meaning that if we find one side, we can simply multiply by four to find the perimeter. Second, the diagonals of a rhombus are perpendicular bisectors of each other, thus giving us four right triangles and splitting each diagonal in half.

Answer 2

Answer:

A rhombus has four sides and all are equal.

i.e. AB = BC = CD = AD

Now, take diagonal AC. This divides the rhombus into two triangles ABC and CDA.

In ΔABC and ΔCDA,

AB = CD {equal sides}

BC = DA {equal sides}

AC = AC {common}

By SSS congruent criterian,

ΔABC ≅ ΔCDA

Now, ΔABC and ΔCDA are isoscales triangle since they have two equal sides

Thus, ∠DAC = ∠DCA = ∠BAC = ∠BCA = DAB/2

Again, take diagonal BD, we can show that,

ΔABD ≅ ΔCBD

and ∠ADB = ∠CDB = ∠ABD = ∠CBD = ABC/2

Let E is the mid point where diagonal AC and BD intersect each other.

So, ∠DAE = ∠BAE = ∠BCE = ∠DCE = ABC/2

and ∠ABE = ∠CBE = ∠CDE = ∠ADE = ABD/2

and AB = BC = CD = AD

Thus, the four triangle are congrent by AAS.

hence, the diagonals of a rhombus divide it into 4 equal congruent triangles.