Question

D
А
M
8. In right triangle ABC, right angled at C, Mis
the mid-point of hypotenuse AB. C is joined
to M and produced to a point D such that
DM = CM. Point D is joined to point B
(see Fig. 7.23). Show that:
(1) A AMC = A BMD
(ii) Z DBC is a right angle.
(iii) A DBC = A ACB
B
C С
Fig. 7.23
1
(iv) CM =
AB
2​

Answer 1

Answer:

I hope it will help...

Step-by-step explanation:

ANSWER

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

DC2=DB2+BC2 (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=21

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