Question

Prove that diagnols of a rhombus devide in four congruent triangle.

Answer 1



Let ABCD is a rhombus.

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.