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 a point D such that DM = CM. Point D is joined to point B as shown in the figure. Show that:
i) ∆AMC ≅ ∆BMD
ii) ∠DBC is a right angle
iii) ∆DBC ≅ ∆ACB

Answer 1

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

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

Answer 2

$\sf{Given :-}$

M is the midpoint of AB

Such that ,

AM = BM

CM = DM

$\sf{To \ prove :-}$

∆AMC ≅ ∆BMD

$\sf{Proof :-}$

In ∆AMC and ∆BMD ,

AM = BM [ Given ]

Angle AMC = Angle BMD [Vertically opposite angles ]

CM = DM

Therefore , ∆AMC ≅ ∆BMD - By S-A-S test

AC = DB - [ By CPCT ] ......(1)

┄───┄───┄───┄───┄───

$\sf{To \ prove :-}$

∠DBC is a right angle

$\sf{Proof :-}$

DBC is a right angled triangle ,

By using Pythagoras theoram ,

DC² = DB² + BC²

So, angle DBC = 90°

┄───┄───┄───┄───┄───

$\sf{To \ prove :-}$

∆DBC ≅ ∆ACB

$\sf{Proof :-}$

BC = BC [ common ]

Angle B = Angle C [ Each 90° ]

AC = DB ......From (1)

Therefore , ∆DBC ≅ ∆ACB - by S-A-S test

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