Question

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Please solve with explanation ​

Answer 1

This figure is symmetrical.

Given

There are 2 right angle triangles in the given figure

so M  is the mid point of hypotenuse AB and DC

Where Triangle ABC=Triangle BCD

So,

1)Triangle AMC=Triangle BMD

2)Angle DBC is a right angle=90 degree

3)Triangle DBC=Triangle ACB

4)CM=I/2 AB because M is the mid point and Hypotenuse of both the triangles are equal.

Answer 2

Working out:

In this question, we are given some conditions and we have to prove the required statements.

We have,

• ABC is a right angled triangle. Right angled at C.
• M is the midpoint of AB as well as DC

$\large{ \underline{ \underline{ \bf{ \red{Proofs}}}}}$

i) △AMC ≅ △BMD

In △AMC and △BMD,

We are given in the question that 'M' is the midpoint of AB, So we can write:

• BM=AM

And also it is the mid point of DC then:

• DM=MC

If we will see for ∠BMD and ∠AMC, then they are equal because they are vertically opposite to each other.

• ∠BMD = ∠AMC

Therefore,

We can say that

∴ △AMC≅△BMD (By SAS congurence)

ii) ∠DBC is a right angle

From (I), We got:

• △AMC≅△BMD

Then,

• ∠ACM = ∠BDM (By CPCT)

But they are also alternate interior angles, and also they are equal. So, we can say that DB || AC

Now, since DB || AC, BC will act as the transversal. ∠ACB = ∠DBC = 90°. Because already given in Q, that angle C = 90°.

Therefore,

We can say that,

∴ ∠DBC = 90°

iii) △DBC≅△ACB

From (I), We got

• AC = BD (By CPCT)

From (II), We got

• ∠DBC = ∠ACB = 90°

And,

• BC = common

Therefore,

We an say that,

∴ △DBC≅△ACB (By SAS congurence)

iv) CM = 1/2 AB

As △DBC≅△ACB ( From III )

• DC = AB (By CPCT)

Now, we can divide 2 in both sides,

• DC / 2 = AB / 2

From (1), We got that,

• DC = CM
• DC = CM = 1/2 AB

Therefore,

We can say that,

∴ CM = 1/2 AB

And, we are done proving all the Q.s !!

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