# 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

Dear Student,

We form our diagram , As : Now from angle sum property of triangle we get in triangle ABP :

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

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

$\angle$ ABP = 25$°$                                                        --- ( 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$°$                                                        --- ( 2 )

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

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

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

$\angle$ QCD = 65$°$                                                                       ( Ans )