Question

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Answer 1

Answer:

∆AMC and BMD

1. AM=BM

2.DM=AM

3.[BMD=AMC] (Adverse Angle)

Answer 2

Answer:

Ans). Given: ∆ABC is a right angled triangle.

<C is 90° .

M is midpoint of hypotenuse AB.

DM=CM

To prove: write your self from question

Proof:. (i) In ∆AMC &∆BMD

DM=CM (given)

AM=BM (M is midpoint)

<AMC=<BMD(vertically opposite)

by side angle side,

∆AMC=∆BMD (proved)

(ii) from part 1,

∆AMC=∆BMD

so, <ACM= <BDM ( CPCT)

But <ACM and <BDM are alternate interior angles for lines AC and BD.

If a transversal intersects two lines such that pair of alternate interior angles is equal, then lines are parallel.

So, BD|| AC

Now, Since

DB || AC and considering BC as transversal

-: <DBC + <ACB= 180°

-: <DBC = 180°-90°

-: <DBC = 90°

Hence,<DBC is a right angle. (proved)

(iii) from part 1,

∆AMC= ∆BMD

AC= BD (CPCT) ......(1)

In ∆DBC & ∆ACB,

DB= AC ( from...(1) )

< DBC= <ACB ( both 90°)

BC= CB ( common)

By side angle side,

Hence, ∆DBC= ∆ACB

(iv) I don't know this one.Sorry

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