A) \[{{90}^{o}}\]
B) \[{{60}^{o}}\]
C) \[{{45}^{o}}\]
D) \[{{135}^{o}}\]
Correct Answer: A
Solution :
As shown in the figure, since P is the midpoint of AB and \[AB=2AD,\]we have \[AB=2AP=2AD\]or \[AP=AD\]. i.e., triangle ADP is an isosceles triangle. If \[\angle ADP={{x}^{o}}\]and \[\angle APD={{x}^{o}},\] then \[\angle A={{180}^{o}}-2{{x}^{o}}\]. Since \[\angle B\] is adjacent to \[\angle A\], in ABCD \[\angle B={{180}^{o}}-({{180}^{o}}-2x)=2x.\] In \[\Delta CBP,\,\,{{x}^{o}}+{{x}^{o}}+2{{x}^{o}}={{180}^{o}}\] (Angle sum property) \[\Rightarrow \] \[4{{x}^{o}}={{180}^{o}}\] \[\Rightarrow \] \[{{x}^{o}}={{45}^{o}}\] \[\therefore \] \[\angle CPD={{180}^{o}}-2{{x}^{o}}\] \[={{180}^{o}}-2\times {{45}^{o}}={{90}^{o}}\]
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