Question

5) The sides AB and CD of a quadrilateral ABCD are extended to points P and Q respectively. Is ∠ADQ + ∠CBP = ∠A + ∠C? Give reason.

Answer 1

Answer:

Join AC, then

∠CBP = ∠BCA + ∠BAC and

∠ADQ = ∠ACD + ∠DAC

(Exterior angles of triangles)

Therefore, ∠CBP + ∠ADQ = ∠BCA + ∠BAC + ∠ACD + ∠DAC

= (∠BCA + ∠ACD) + (∠BAC + ∠DAC)

= ∠C + ∠A

Hope it helps!

Answer 2

Yes, in quadrilateral ABCD , ∠ADQ + ∠CBP = ∠A + ∠C. It is proved below by the exterior angle property of a triangle.

Given:

Sides AB and CD of a quadrilateral ABCD are extended to points P and Q.

To prove :

∠ADQ + ∠CBP = ∠A + ∠C

Concept used :

Exterior angle property of a triangle :

"In a triangle, if any of its sides is produced, then the exterior angle formed by producing the side to the triangle is equal to the sum of its two opposite interior angles."

For e.g

In ∆ABC side, BC is produced to D. Then ∠ACD = ∠ABC + ∠BAC

Fig. is in the attachment below.

Solution:

Step 1  of 2:

Draw a quadrilateral ABCD with the data given and join AC. Fig. is in the attachment below.

Using the Exterior angle property of a triangle in ∆ABC and ∆ADC:

In ∆ADC,  ∠ADQ is an exterior angle:

∠ADQ = ∠ACD + ∠DAC ......(1)

In ∆ABC, ∠CBP is an exterior angle:

∠CBP = ∠BCA + ∠BAC .......(2)

Step 2  of 2:

Adding eq. 1 and 2, we have :

∠ADQ + ∠CBP =  (∠ACD + ∠DAC) + (∠BCA + ∠BAC)

∠ADQ + ∠CBP =  (∠BAC + ∠DAC) + (∠BCA + ∠ACD)

∠ADQ + ∠CBP = ∠A + ∠C

Hence it is proved that , ∠ADQ + ∠CBP = ∠A + ∠C

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