Question-217256

Facebook Tweet Pin

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

further we can also find AC=BD and AC⊥BD

$${further}\:{we}\:{can}\:{also}\:{find} \\ $$$${AC}={BD}\:{and}\:{AC}\bot{BD} \\ $$

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

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

∠ABC=∠BCD=π−(π/4)−((π−θ)/2)=(π/4)+(θ/2) ∠BAD=∠ADC=π−(π/4)−(θ/2)=((3π)/4)−(θ/2) ∠BAD+∠ABC=∠ADC+∠BCD=π AD//BC ⇒ABCD is isosceles trapezoid ⇒AB=CD ⇒(√2)r_1 +(√2)r_2 =(√2)r_3 +(√2)r_4 ⇒r_1 +r_2 =r_3 +r_4 ✓

$$\angle{ABC}=\angle{BCD}=\pi−\frac{\pi}{\mathrm{4}}−\frac{\pi−\theta}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}} \\ $$$$\angle{BAD}=\angle{ADC}=\pi−\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}=\frac{\mathrm{3}\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}} \\ $$$$\angle{BAD}+\angle{ABC}=\angle{ADC}+\angle{BCD}=\pi \\ $$$${AD}//{BC} \\ $$$$\Rightarrow{ABCD}\:{is}\:{isosceles}\:{trapezoid} \\ $$$$\Rightarrow{AB}={CD} \\ $$$$\Rightarrow\sqrt{\mathrm{2}}{r}_{\mathrm{1}} +\sqrt{\mathrm{2}}{r}_{\mathrm{2}} =\sqrt{\mathrm{2}}{r}_{\mathrm{3}} +\sqrt{\mathrm{2}}{r}_{\mathrm{4}} \\ $$$$\Rightarrow{r}_{\mathrm{1}} +{r}_{\mathrm{2}} ={r}_{\mathrm{3}} +{r}_{\mathrm{4}} \:\:\:\checkmark \\ $$

Commented by mnjuly1970 last updated on 08/Mar/25