Giải thích các bước giải:

a.Xét $\Delta PQR, \Delta HPQ$ có:
Chung $\hat Q$

$\widehat{PHQ}=\widehat{QPR}(=90^o)$

$\to \Delta PQR\sim\Delta HQP(g.g)$

$\to \dfrac{PR}{PH}=\dfrac{QR}{QP}$

$\to PH=\dfrac{PR\cdot QP}{QR}$

Vì $\Delta PQR$ vuông tại $P\to PR=\sqrt{QR^2-QP^2}=3$

$\to PH=\dfrac{12}{5}$

b.Xét $\Delta PHQ, \Delta PHR$ có:

$\widehat{PHQ}=\widehat{PHR}(=90^o)$

$\widehat{QPH}=90^o-\widehat{HPR}=\widehat{PRH}$

$\to \Delta PHQ\sim\Delta RHP(g.g)$

$\to \dfrac{PH}{RH}=\dfrac{HQ}{HP}$

$\to HP^2=HR.HQ$

c.Ta có $RD$ là phân giác $\widehat{PRQ}$

$\to \dfrac{DQ}{DP}=\dfrac{RQ}{RP}=\dfrac54$

$\to \dfrac{DQ}{DQ+DP}=\dfrac5{5+4}$

$\to \dfrac{DQ}{PQ}=\dfrac59$

$\to\dfrac{S_{RDQ}}{S_{PQR}}=\dfrac59$

Mà $S_{PRQ}=\dfrac12PR.PQ=6$

$\to S_{RDQ}=\dfrac{10}{3}$

Ta có $HQ=\sqrt{PQ^2-PH^2}=\dfrac95\to HR=QR-HQ=\dfrac{16}{5}$

$\to S_{DRH}=\dfrac{HR}{QR}S_{DQR}=\dfrac{16}{25}S_{DQR}=\dfrac{32}{15}$

d.Xét $\Delta PHK, \Delta PQH$ có:

chung $\hat P$

$\widehat{PKH}=\widehat{PHQ}(=90^o)$

$\to\Delta PKH\sim\Delta PHQ(g.g)$

$\to \dfrac{PK}{PH}=\dfrac{PH}{PQ}$

$\to PH^2=PK.PQ$

Tương tự $PH^2=PI.PR$

$\to PK.PQ=PI.PR$