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Question Number 217256 by mnjuly1970 last updated on 07/Mar/25

Answered by mr W last updated on 08/Mar/25

Commented by mr W last updated on 08/Mar/25

$$furtherwecanalsofind$$$$AC=BDandAC\mathrm{\perp }BD$$

Commented by mr W last updated on 08/Mar/25

Commented by mr W last updated on 08/Mar/25

$$\mathrm{\angle }ABC=\mathrm{\angle }BCD=\pi -\frac{\pi }{4}-\frac{\pi -\theta }{2}=\frac{\pi }{4}+\frac{\theta }{2}$$$$\mathrm{\angle }BAD=\mathrm{\angle }ADC=\pi -\frac{\pi }{4}-\frac{\theta }{2}=\frac{3\pi }{4}-\frac{\theta }{2}$$$$\mathrm{\angle }BAD+\mathrm{\angle }ABC=\mathrm{\angle }ADC+\mathrm{\angle }BCD=\pi$$$$AD//BC$$$$\Rightarrow ABCDisisoscelestrapezoid$$$$\Rightarrow AB=CD$$$$\Rightarrow \sqrt{2}{r}_1+\sqrt{2}{r}_2=\sqrt{2}{r}_3+\sqrt{2}{r}_4$$$$\Rightarrow r_1+r_2=r_3+r_4✓$$

Commented by mnjuly1970 last updated on 08/Mar/25

$$⋚$$

Commented by mnjuly1970 last updated on 08/Mar/25

$$\underset{―}{\underbrace{}}$$

Commented by mr W last updated on 23/Mar/25

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