Успоредници

[Jdoe]

Някой може ли да ми обясни как се получават отговорите?

[liam]

Някой може ли да ми обясни как се получават отговорите?

[ammornil]

$ABCD, \quad AB=CD, \quad AD=BC, \quad \measuredangle{ABC}= \measuredangle{ADC}= \measuredangle{BAD}= 90^{\circ}\\[6pt] M\in{CD}, \quad AM=CD\\[12pt] \text{(А)}: \\ \begin{cases}AB=CD \\ AM=CD \end{cases} \Rightarrow AB=AM \Rightarrow \measuredangle{AMB}= \measuredangle{ABM} =\varphi \\[6pt] \measuredangle{MBC}= 90^{\circ}- \measuredangle{ABM}= 90^{\circ} -\varphi \\[6pt] \triangle{BCM}: \quad \measuredangle{BMC}= 90^{\circ}- \measuredangle{MBC}= 90^{\circ}-(90^{\circ}-\varphi)= \varphi \\[6pt] \measuredangle{AMB}= \measuredangle{BMC} \Rightarrow MB \text{ е ъглополовяща на } \measuredangle{AMC} \\[24pt] \text{(Б)}:\\AB=12, \quad S_{ABCD}= AB\cdot{BC} \Rightarrow BC=\dfrac{S_{ABCD}}{AB}=8 \\[6pt] P\in{AM}, \quad BP\bot{AM} \\[6pt] \triangle{BPM}: \quad \angle{PBM}= 90^{\circ}-\varphi \\[6pt] \triangle{MPB}\cong\triangle{MCB} \begin{cases} PB \text{ е обща} \\ \measuredangle{PMB}= \measuredangle{CMB} \\ \measuredangle{PBM}= \measuredangle{CBM} \end{cases} \rightarrow \text{ II признак } \Rightarrow BP=BC= 8\\[12pt]$

Screenshot 2025-05-02 183040.png (23.38 KiB) Прегледано 518 пъти