làm bài trên ảnh nhé

Giải thích các bước giải:

a.Xét $\Delta AHB,\Delta CDA$ có:

$\widehat{AHB}=\widehat{ADC}(=90^o)$

$\widehat{HAB}=\widehat{ACD}$ vì $AB//CD$

$\to \Delta AHB\sim\Delta CDA(g.g)$

b.Xét $\Delta BCM,\Delta ADC$ có:

$\widehat{BCM}=\widehat{ADC}(=90^o)$

$\widehat{MBC}=90^o-\widehat{HCB}=\widehat{ACD}$

$\to \Delta BCM\sim\Delta CDA(g.g)$

$\to \dfrac{BC}{CD}=\dfrac{CM}{DA}$

$\to CM\cdot CD=BC\cdot DA$

Ta có: $CD=AB=8, AD=BC=6$

$\to CM=\dfrac{BC\cdot AD}{CD}=\dfrac92$

$\to S_{BMC}=\dfrac12CM\cdot BC=\dfrac12\cdot\dfrac92\cdot 6=\dfrac{27}2$

c.Xét $\Delta MIC,\Delta BMK$ có:

$\widehat{IMC}=\widehat{MKB}(=90^o)$

$\widehat{ICM}=90^o-\widehat{MIH}=\widehat{IMH}=\widehat{KMB}$

$\to \Delta MIC\sim\Delta KBM(g.g)$

$\to \dfrac{MI}{KB}=\dfrac{IC}{MB}$

$\to MI\cdot BM=KB\cdot IC$