Giải bài 48 giải ngắn gọn

Giải thích các bước giải:

a.Ta có:

$\Delta TMC\sim\Delta TNB(g.g)$

$\to \dfrac{TM}{TN}=\dfrac{TC}{TB}$

$\to \Delta TMN\sim\Delta TCB(c.g.c)$

$\to \widehat{TMN}=\widehat{TCB}$

$\to \widehat{BMN}=\widehat{TCB}$

Ta có:

$\Delta BHT\sim\Delta BMC(g.g)$

$\to \dfrac{BH}{BM}=\dfrac{BT}{BC}$

$\to \Delta BHM\sim\Delta BTC(c.g.c)$

$\to \widehat{BMH}=\widehat{BCT}=\widehat{BMN}$

$\to \widehat{BMN}=\widehat{TMH}$

Ta có: $\Delta TAM\sim\Delta TBH(g.g)$

$\to \dfrac{TA}{TB}=\dfrac{TM}{HT}$

$\to \Delta TAB\sim\Delta TMH(c.g.c)$

$\to \widehat{TBA}=\widehat{THM}$

$\to \widehat{NBM}=\widehat{THM}$

$\to \Delta BMN\sim\Delta HMT(g.g)$

$\to \dfrac{MB}{MH}=\dfrac{MN}{MT}$

$\to MH.MN=MB.MT$

b.Ta có: $TQ\cap C=I$ là trung điểm mỗi đường

$\to BTCQ$ là hình bình hành
$\to TB=CQ, CT=BQ$

Ta có: $\Delta HBT\sim\Delta HAC(g.g)$

$\to \dfrac{BT}{AC}=\dfrac{HB}{HA}$

$\to \dfrac{CQ}{AC}=\dfrac{HB}{HA}$

Lại có:

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

$\to \Delta AHB\sim\Delta ACQ(c.g.c)$

$\to \widehat{QAC}=\widehat{HAB}$

Gọi $AQ\cap MN=D$

$\to \widehat{DAM}=\widehat{HAB}$

Mà $\Delta AMB\sim\Delta ANC(g.g)$

$\to \dfrac{AM}{AN}=\dfrac{AB}{AC}$

$\to \Delta AMN\sim\Delta ABC(c.g.c)$

$\to \widehat{AMN}=\widehat{ABC}$

$\to \widehat{AMD}=\widehat{ABH}$

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

$\to \widehat{ADM}=\widehat{AHB}=90^o$

$\to AQ\perp MN$

c.Từ b $\to \widehat{ANM}=\widehat{ACB}$

$\to \widehat{KNB}=\widehat{ANM}=\widehat{KCM}$

$\to \Delta KBN\sim\Delta KMC(g.g)$

$\to \dfrac{KB}{KM}=\dfrac{KN}{KC}$

$\to KB.KC=KM.KN$

Tương tự a chứng minh được $\widehat{MNC}=\widehat{CNH}=\widehat{MBC}$

Ta có: $\Delta BMC$ vuông tại $M, I$ là trung điểm $BC$

$\to IM=IB=IC=\dfrac12BC$

$\to \Delta IMC$ cân tại $I$

$\to \widehat{MIC}=2\widehat{IBM}=2\widehat{MNC}=\widehat{MNH}$

$\to \Delta KNH\sim\Delta KIM(g.g)$

 $\to \dfrac{KN}{KI}=\dfrac{KH}{KM}$

$\to \Delta KNI\sim\Delta KHM(c.g.c)$

$\to \widehat{KHM}=\widehat{KNI}$

$\to \widehat{IHM}=180^o-\widehat{KHM}=180^o-\widehat{KNI}=\widehat{INM}=\widehat{IMN}$ vì $IM=IN=IB=IC=\dfrac12BC$

$\to \Delta IHM\sim\Delta IMK(g.g)$

$\to \dfrac{IH}{IM}=\dfrac{IM}{IK}$

$\to IH.IK=IM^2$

$\to IH.IK=IC^2$

$\to IH.(IH+HK)=IC^2$

$\to HK.HI+IH^2=IC^2$

$\to HK.HI=IC^2-IH^2$

$\to HK.HI=(IC-IH)(IC+IH)$

$\to HK.HI=(IB-IH)(IC+IH)$

$\to HK.HI=HB.HC$

$\to HK.HI=HA.HT$

$\to \dfrac{HK}{HA}=\dfrac{HT}{HI}$

$\to \Delta HKT\sim\Delta HAI(c.g.c)$

$\to\widehat{KTH}=\widehat{HIA}$

$\to KT\perp AI$

Mà $AT\perp IK$

$\to T$ là trực tâm $\Delta AKI$

$\to IT\perp AK$