Giải thích các bước giải:
a.Xét $\Delta ABC,\Delta AHC$ có:
Chung $\hat C$
$\widehat{AHC}=\widehat{CAB}(=90^o)$
$\to \Delta ABC\sim\Delta HAC(g.g)$
b.Xét $\Delta AHB,\Delta ACH$ có:
$\widehat{AHB}=\widehat{AHC}(=90^o)$
$\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}$
$\to \Delta AHB\sim\Delta CHA(g.g)$
$\to \dfrac{AH}{CH}=\dfrac{HB}{HA}$
$\to AH^2=HB.HC$
c.Ta có $BC=HB+HC=25$
$AH^2=HB.HC=144\to AH=12$
$\to AB=\sqrt{AH^2+HB^2}=15, AC=\sqrt{AH^2+HC^2}=20$
Ta có $BE$ là phân giác $\hat B$
$\to \dfrac{EA}{EC}=\dfrac{BA}{BC}=\dfrac35$
$\to \dfrac{EA}{EA+EC}=\dfrac3{3+5}$
$\to \dfrac{EA}{AC}=\dfrac38$
$\to AE=\dfrac38AC=7.5$
$\to EC=AC-AE=12.5$
d.Ta có $EM, EN$ là phân giác $\widehat{AEB},\widehat{BEC}$ khi đó:
$\dfrac{MB}{MA}.\dfrac{AE}{EC}.\dfrac{NC}{NB}$
$=\dfrac{EB}{EA}.\dfrac{AE}{EC}.\dfrac{EC}{EB}$
$=1$
$\to đpcm$