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Question Number 177044 by Ar Brandon last updated on 30/Sep/22

Answered by a.lgnaoui last updated on 30/Sep/22

$$\mathrm{\Sigma }anglesinterieures=\mathrm{\Sigma }anglesexterieures$$$$\frac{\measuredangle B}{2}+\measuredangle KPD+\frac{\measuredangle D}{2}=90$$$$\measuredangle EDD=\frac{\measuredangle D}{2}(PK\mid \mid ED)$$$$\frac{\measuredangle B}{2}+\measuredangle D=90$$$$quadrilateralAPKB={360}^{°}$$$$\frac{B}{2}=360-180-160={20}^{°}$$$$donc\measuredangle D=90-20=70$$$$$$$$m\left(PDC\right)=\frac{\measuredangle D}{2}={35}^{°}$$

Commented by Ar Brandon last updated on 30/Sep/22

Merci monsieur

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