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Question Number 113063 by Aina Samuel Temidayo last updated on 11/Sep/20

Answered by nimnim last updated on 11/Sep/20

Commented by nimnim last updated on 11/Sep/20

$$\alpha =\alpha \left(\mathrm{angles}\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{segment}\right)$$$$\gamma =90+\alpha (\mathrm{DLQ}\mathrm{is}\mathrm{rt}.△)$$$$\alpha +\beta =90(\mathrm{AMB}\mathrm{is}\mathrm{rt}.△)$$$$\mathrm{in}\mathrm{quad}.\mathrm{ABQP},$$$$\mathrm{\angle }\mathrm{BAP}+\mathrm{\angle }\mathrm{BQP}=\beta +(90+\alpha )=180$$

Commented by Aina Samuel Temidayo last updated on 11/Sep/20

$$\mathrm{So}\mathrm{the}\mathrm{proof}\mathrm{is}\mathrm{this}\mathrm{short}?\mathrm{Thanks}.$$

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