Question 5

In the parallelogram $$ABCD$$, the diagonal $$AC$$ is perpendicular to the side $$AD$$. $$AH$$ is drawn perpendicular to $$CD$$. The tangent at $$D$$ to the circumcircle of $$\triangle ADB$$ meets $$CA$$ produced at $$P$$. Prove that $$\angle PBA = \angle DBH$$.


Correct Answer: 0

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