Question

Prove that the diagonals of a rectangle divide it in two congruent triangles.

Answer 1

Answer:

Given - A rectangle ABCD and BD & AC are the diagonals

Prove - ΔABC ≅ ΔCDA & ΔBCD≅ ΔBAD

Proof - In ΔABC & ΔCDA

AB = CD ( sides of a rect.)

BC = DA (sides of a rect.)

AC = CA (common side)

∴ΔABC ≅ ΔCDA (SSS ≅)

In ΔBAD & ΔBCD

AB = CD ( sides of a rect.)

BC = DA (sides of a rect.)

BD = BD (common side)

∴ ΔBAD ≅ ΔBCD ( SSS≅)

∴ The diagonals of a rect. divide it into 2 congruent triangles ( Sorry can't draw the rect ABCD  and AC & BD the diagonals)

Answer 2

Let, the rectangle be ABCD.

Here, BD and AC are diagonals of rectangle ABCD.

To prove:-

• ∆ABC ≌ ∆ADC
• ∆ABD ≌ ∆DCB

Proof :

Given that,

ABCD is a rectangle.

So,

∠A = ∠B = ∠C = ∠D = 90° [Because all angles of rectangle of 90°] -------(i)

And,

AB = DC and BC = AD [Because opposite sides of

rectangle are equal and parallel] -------(ii)

Now,

According to diagonal AC.

In ∆ABC and ∆CDA

AB = CD [By equation (ii)]

∠B = ∠D [By equation (i) ]

BC = DA [By equation (ii) ]

By SAS congruency

∆ABC ≌ ∆CDA

According to diagonal BD.

In ∆ABD and ∆DCB

AB = CD [By equation (ii)]

∠A = ∠C [By equation (i) ]

BC = DA [By equation (ii) ]

By SAS congruency

∆ABD ≌ ∆DCB

Hence, Proved!!

That,

The Diagonals of rectangle divides the rectangle in two congruent triangles.