Question

ABC is a triangle right angled at C. A line through the mid point of hypothesis AB and parallel to BC intersects AC at D. Show that
i) D is the mid point of AC
ii) MD is perpendicular to AC
iii) CM=MA=½AB ​

Answer 1

Answer:

In ∆ABC, we have

M is the midpoint of AB and MD||BC

D is the midpoint of AC {by converse of midpoint theorem}

Now

MP||BC

<MDC+<BCD=180°

<MDC+90°=180°

<MDC=90°

Thus;MD perpendicular AC

Join MC

In ∆MDA and ∆MDC we have

DA=DC

<MDA=<CDM

MD=MD

∆MDA~∆MDC {S.A.S}

And so MA=MC

Now M is the midpoint of AB

MA=MC=1/2AB

Step-by-step explanation:

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