# In a parallelogram ABCD, AP and CQ are drawn perpendicular to the diogonal BD on measuring it is found that angle PAB =65 and angleDAB =75 find the measure of angle QCD

Please find below the solution to the asked query:

We form our diagram , As :

Now from angle sum property of triangle we get in triangle ABP :

$\angle $ PAB + $\angle $ APB + $\angle $ ABP = 180$\xb0$ , NOw we substitute values from above diagram we get :

65$\xb0$ + 90$\xb0$ + $\angle $ ABP = 180$\xb0$ ,

$\angle $ ABP = 25$\xb0$ --- ( 1 )

And

$\angle $ ABP = $\angle $ CDQ ( Alternate interior angles as AB | | CD and BD is transversal line )

From equation 1 we get :

$\angle $ CDQ = 25$\xb0$ --- ( 2 )

Now from angle sum property of triangle we get in triangle CDQ :

$\angle $ CDQ + $\angle $ CQD + $\angle $ QCD= 180$\xb0$ , Now we substitute values from above diagram and from equation 2 we get :

25$\xb0$ + 90$\xb0$ + $\angle $ QCD = 180$\xb0$ ,

**$\angle $ QCD = 65$\xb0$ ( Ans )**

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