Question

In a rhombus ABCD,AB=BD then what is the measure of angle acd and angle ADC respectively ​

Answer 1

We have a rhombus $ABCD$, with$AB = BD$

Since all sides of a rhombus are equal, so we get

$AD=CD=BC=AB=BD$

Hence, △$ABD$ and △$BCD$, △

So, we get all their angles are equal to $60^o$

So, ∠$ADB=60^o$ and ∠$CDB=60^o$

Therefore, ∠$ADC=$∠$ADB+$∠$CDB=120^o$

Now, consider △$ADC$

As we have $AD=CD$,

So ∠∠      ....(Angles opposite to equal sides are equal)

⇒∠∠∠     ....(by angle sum property)

⇒∠$ADC+2$ ∠$DCA=180^o$

(since ∠$DCA=$∠$DAC$)

⇒$2$∠$DCA=180^o-$∠$ADC=180^o-120^o$

⇒2∠$DCA=60^o$

⇒∠$DCA=30^o$

Hence proved.

Answer 2

As we know that,

Given:

In a rhombus, ABCD, AB=BD  

Find out:

The measure of angle ACD and angle ADC.

Let us:

as all sides of a rhombus are equal, so we get,

AD = CD = BC = BD

So,

ΔABD and ΔBCD,

we get all their angles are equal to 60°

∠ADB=60° and ∠CDB=60°

Therefore, ∠ADC=∠ADB=∠CDB=120°

consider △ADC,

AD=CD

The angles opposite to equal sides are equal angles.

∠∠∠(by angle sum property)

∠ ADC+2∠DCA=180°

(since ∠DCA=∠DAC)

2∠DCA=180°-∠AD180°-120°

⇒2∠DCA=60°

⇒∠DCA=30°

By the angle sum property, it is proved.