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Can someone please help !

Can someone please help !-example-1
User Lukbl
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\huge\begin{array}{ccc}BT=21\\DC=10.2\end{array}

If BT is the midsegment of ΔMAH and DC is the midsegment of ΔBTH, then MA, BT and DC are parallel.

Therefore the segments form the proportion:


(HM)/(MA)=(HB)/(BT)=(HD)/(DC)

BT is the midsegment of ΔMAH, therefore


HB=(1)/(2)HM

DC is the midsegment of ΔBTH, therefore


HD=(1)/(2)HB=(1)/(2)\cdot(1)/(2)HM=(1)/(4)HM

Substitute:


(HM)/(42)=((1)/(2)HM)/(BT)

cross multiply


HM\cdot BT=42\cdot(1)/(2)HM

divide both sides by HM


\boxed{BT=21}


(HM)/(42)=((1)/(4)HM)/(DC)

cross multiply


HM\cdot DC=42\cdot(1)/(4)HM

divide both sides by HM


\boxed{DC=10.5}

User Dercz
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