Question

⚠️ Don't give unusual answers!!!​

Answer 1

) △AMC≅△BMD

Proof: As 'M' is the midpoint

BM=AM

And also it is the mid point of DC then

DM=MC

And AC=DB (same length)

∴Therefore we can say that

∴△AMC≅△BMD

ii) ∠DBC is a right angle

As △DBC is a right angle triangle and

DC 2 =DB 2 +BC 2 (Pythagoras)

So, ∠B=90°

∴∠DBC is 90°

iii) △DBC≅△ACB

As M is the midpoint of AB and DC. So, DM=MC and AB=BM

∴DC=AB (As they are in same length)

And also, AC=DB

and ∠B=∠C=90°

By SAS Axiom

∴△DBC≅△ACB

iv) CM= 1/2 AB

As △DBC≅△ACB

CM=2/DC

∴DC=AB(△DBC≅△ACB)

So, CM= AB/2

∴CM= 1/2AB

u r also kvian !!

I know this question because this question is from class 9 cbse NCERT Book

I also studied !

Answer 2

Answer:

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