Question

Prove that"the diagonal of a parallelogram divides it into 2 congruent triangles"

Answer 1

Answer:

Let ABCD be a parallelogram with diagonal DB..

To prove :∆ADB Congruent ∆BDC...

proof:---

in triangle ADB and ∆BDC..

AD =BC(given)

BD=BD(common)

AB=DC(given)

:. triangle ADB CONGRUENT triangle BDC...

(hence proved)...

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Answer 2

$\huge{\underline{\underline{\green{\mathsf{Answer:}}}}}$

Refer to the attachment

Given: ABCD is a parallelogram

So, AB || CD and AD || BC

In ∆ABC and ∆ADC

<BAC = <ACD ( Alternate interior angles)

<DAC = <ACB ( Alternate interior angles)

AC = AC (Common side)

So, ∆ABC ≅ ∆ADC

{By AAS congruency criteria}

From this we can say that the diagonal of a parallelogram divides it into two congruent triangles