$\color{darkgoldenrod}{a)}\ \ \ $Xét $\triangle DAH$ và $\triangle HAC$ có:

$\hspace{0.65cm}\ \widehat{HAC}$ chung

$\hspace{0.65cm}\ \widehat{ADH}=\widehat{AHC}\ (=90^\circ )$

$\Rightarrow\ \triangle DAH\ \backsim\ \triangle HAC\ (g-g)\ (đpcm)$

$\\$$\\$

$\color{darkgoldenrod}{b)}\ \ \ AB\bot AC$

$\hspace{0.65cm}\ HD\bot AC$

$\Rightarrow\ HD//AB$

$\Rightarrow\ \dfrac{IH}{OB}=\dfrac{IC}{OC}=\dfrac{ID}{OA}\ (Thales)$

$\hspace{0.65cm}\ $Mà $OA=OB\ (O$ là trung điểm $AB)$

$\Rightarrow\ IH=ID\ \ (đpcm)$

$\\$$\\$

$\color{darkgoldenrod}{c)}\ \ \ \triangle ABC$ vuông tại $A$ có $AH$ là đường cao

$\Rightarrow\ BH.BC=AH^2\ \ (htlttgv)\ (1)$

$\hspace{0.65cm}\ \triangle HAC$ vuông tại $H$ có $HD$ là đường cao

$\Rightarrow\ AD.AC=AH^2\ \ (htlttgv)\ (2)$

$\hspace{0.65cm}\ $Từ $(1),(2)\ \Rightarrow\ AD.AC=BH.HC\ \ (đpcm)$

$\\$$\\$

$\color{darkgoldenrod}{d)}\ \ \ $Xét $\triangle HKI$ và $\triangle AKO$ có:

$\hspace{0.65cm}\ \widehat{HKI}=\widehat{AKO}$ (đối đỉnh)

$\hspace{0.65cm}\ \widehat{KHI}=\widehat{KAO}$ (so le trong)

$\Rightarrow\ \triangle HKI\ \backsim\ \triangle AKO\ (g-g)$

$\Rightarrow\ \dfrac{IK}{KO}=\dfrac{IH}{OA}=\dfrac{ID}{OB}$

$\hspace{0.65cm}\ $Xét $\triangle IKD$ và $\triangle OKB$ có:

$\hspace{0.65cm}\ \dfrac{IK}{KO}=\dfrac{ID}{OB}\ \ (cmt)$

$\hspace{0.65cm}\ \widehat{KID}=\widehat{KOB}$ (so le trong)

$\Rightarrow\ \triangle IKD\ \backsim\ \triangle OKB\ (c-g-c)$

$\Rightarrow\ \widehat{IKD}=\widehat{OKB}$

$\Rightarrow\ \overline{B,K,D}\ \ (đpcm)$

Đáp án:

Giải thích các bước giải:

 Xin CTLHN