Question

Please answer this as fast as possible....​

Answer 1

GIVEN :-

• The sides AB and CD of a quadrilateral ABCD are extended to points P and Q.

TO PROVE :-

• ∠ADQ + ∠CBP = ∠A + ∠C.

SOLUTION :-

✮ In ∆ ABC By Exterior angle property,

⇒ ∠CBP = ∠BCA + ∠BAC. ...Eq(1).

✮ In ∆ ACD By Exterior angle property,

⇒ ∠ADQ = ∠DCA + ∠DAC. ...Eq(2).

✮ Adding Equation (1) and (2),

⇒ ∠CBP +∠ ADQ = ∠BCA + ∠BAC + ∠DCA + ∠DAC.

From the given figure we have,

⇒ ∠C = ∠DCA + ∠BCA

⇒ ∠A = ∠DAC + ∠BAC

Now,

⇒ ∠CBP +∠ ADQ = ∠BCA + ∠BAC + ∠DCA + ∠DAC.

⇒ ∠CBP +∠ ADQ = (∠DAC + ∠BAC) + (∠DCA + ∠BCA)

⇒ ∠CBP +∠ ADQ = ∠A + ∠C.

Hence Proved.

Answer 2

Given :

• ABCD are extended to points P and Q

To find :

• ∠ADQ + ∠CBP = ∠A + ∠C

According to the question :

Joining AC, then we get

➳ ∠CBP = ∠BCA + ∠BAC and

➳ ∠ADQ = ∠ACD + ∠DAC

( Exterior angles of Triangles )

Equation will be :

⟹ ∴ ∠CBP + ∠ADQ = ∠BCA + ∠BAC + ∠ACD + ∠DAC

⟹ ( ∠BCA + ∠ACD ) + ( ∠BAC + ∠DAC )

⟹ ∴ ∠C + ∠A

Hence Proved !

Note :

Refer to the picture

So, It's Done !!