Question

I will mark as brainiest if anyone answers.

Answer 1

1 is 40

2 is 70

3 is 60

4 is 90

mark me brainiest

Answer 2

Step-by-step explanation:

It is given that in parallelogram ABCD, BP is perpendicular to AC and DQ is perpendicular to AC.

In ΔADQ and ΔCBP,

AD=CD  (Opposite sides of a parallelogram)

∠DAQ=∠BCP and AD∣∣BC,AC  (Transversal alternate angles)

∠DQA=∠BPC=900  (Given)

⸫ΔADQ=ΔCBP         (SAA)

⸫BP=DQ                    (C.P.C.T)

Hence proved