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
$$\alpha=\alpha\left(\mathrm{angles}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{segment}\right) \\ $$$$\gamma=\mathrm{90}+\alpha\left(\mathrm{DLQ}\:\mathrm{is}\:\mathrm{rt}.\:\bigtriangleup\right) \\ $$$$\alpha+\beta=\mathrm{90}\left(\mathrm{AMB}\:\mathrm{is}\:\mathrm{rt}.\:\bigtriangleup\right) \\ $$$$\mathrm{in}\:\mathrm{quad}.\mathrm{ABQP}, \\ $$$$\angle\mathrm{BAP}+\angle\mathrm{BQP}=\beta+\left(\mathrm{90}+\alpha\right)=\mathrm{180} \\ $$
Commented by Aina Samuel Temidayo last updated on 11/Sep/20
$$\mathrm{So}\:\mathrm{the}\:\mathrm{proof}\:\mathrm{is}\:\mathrm{this}\:\mathrm{short}?\:\mathrm{Thanks}. \\ $$