Exercițiul 760

E.760. Dacă A^=D^=90°, P∈(AD), CM=CA, BN=BA,\widehat{A}=\widehat{D}=90\degree,~P \in (AD),~CM=CA,~BN=BA, atunci MQ=NQ.MQ=NQ.

Alin Crețu (trunchiată de LM)
Soluție
Soluție:

Fie CF⊥BPCF \perp BP și CF∩DP={H}.CF \cap DP = \{H\}. Rezultă CP⊥BH.CP \perp BH.

MC2=AC2=CD⋅CB⇒MCCD=CBMC⇒△CBM∼△CMD⇒B1^=M1^.MC^2=AC^2=CD \cdot CB \Rightarrow \dfrac{MC}{CD} = \dfrac{CB}{MC} \Rightarrow \triangle CBM \sim \triangle CMD \Rightarrow \boxed{\widehat{B_1} = \widehat{M_1}}.

D^=F^=90°⇒BDFH\widehat{D} = \widehat{F}=90\degree \Rightarrow BDFH inscriptibil ⇒B1^=H1^.\Rightarrow \boxed{\widehat{B_1} = \widehat{H_1}}.

Deci M1^=H1^⇒CDMH\widehat{M_1} = \widehat{H_1} \Rightarrow CDMH inscriptibil ⇒CMH^=90°.\Rightarrow \boxed{\widehat{CMH}=90\degree}. Analog, BNH^=90°.\boxed{\widehat{BNH}=90\degree}.

HM2=HF⋅HC△HFP∼△HDC⇒HF⋅HC=HP⋅HD}⇒HM2=HP⋅HD. \begin{rcases} HM^2=HF \cdot HC \\ \triangle HFP \sim \triangle HDC \Rightarrow HF \cdot HC = HP \cdot HD \end{rcases} \Rightarrow \boxed{HM^2=HP \cdot HD}.

Analog, HN2=HP⋅HD.\boxed{HN^2=HP \cdot HD}. Deci HM=HN.\boxed{HM=HN}.

△HMQ≡△HNQ (I.C.)⇒MQ=NQ.\triangle HMQ \equiv \triangle HNQ~ (I.C.) \Rightarrow \boxed{MQ=NQ}.