Question

can anyone help me to solve​

Answer 1

$\huge\rm\underline\purple{Given :-}$

• In a triangle ABC right angled at C
• M is the mid-point kn hypotenuse AB
• DM = CM

$\huge\rm\underline\purple{To Show :-}$

• △ AMC ≅ △ BMD
• ∠DBC is right angled
• △DBC ≅ △ACB
• CM = 1/2 AB

$\huge\rm\underline\purple{Solution :-}$

(i) As M is the mid point

BM = AM

And also it is the mid point on DC

then, DM = MC

And AC = BD ( same length)

Therefore, we can say that

△ AMC ≅ △ BMD

(ii) ∠DBC is right angled

As ∠DBC is right angled triangle and

DC² = DB² + BC² (Pythagoras)

so, ∠B = 90°

∠DBC = 90°

(iii) △DBC ≅ △ACB

As M is the mid-point on AB and DC.

So, DM = MC AND AB = BM

DC = AB (same length)

and also DC = DB

and ∠B = ∠C = 90°

By SAS criteria

△DBC ≅ △ACB

(iv) CM = 1/2 AB

As △DBC ≅ △ACB

CM = DC/2

DC = AB (△DBC ≅ △ACB)

So, CM = AB/2

CM = 1/2AB

HOPE IT HELPS

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