Question

пж 3 и 4 задачу прошууу

Answer 1

3. $alpha + beta + gamma =180^o$
$alpha_{2} + beta_{2} + gamma_{2} = 180^o$
$alpha = alpha_{2}=90^o$
${ beta +gamma=180^o-alpha=90$
$left { {{gamma=90^o-75^o} atop {beta= 75^o}} right. left { {{beta_{2} =90^o-15^o} atop {gamma_{2}=15^o}} right.$
$beta=beta_{2}$
$gamma=gamma_{2}$
$зBDE=зKMC$
По трем углам.