Question

.In right triangle ∆ABC right angled at C, M is the mid point of
hypotenuse AB, C is joined to M and produced to point D such that
DM=CM. Point D is joined to point B. Show that,
i. ∆ AMC ≅ ∆BMD
ii. ∟DBC = 90°
iii. ∆ DBC ≅ ∆ ACB
v. CM = 1/2 AB
class 9

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

i) △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=

2

1

AB

As △DBC≅△ACB

CM=

2

DC

∴DC=AB(△DBC≅△ACB)

So, CM=

2

AB

∴CM=

2

1

AB

solution