Question

6. ABC is a triangle right angled at c. line that
mid points of hypotenuse A and
parallel to BC intersect AC at O.
Show that ..
1) D mid point of AC
2)MD perpendivular toAC
3) CM = MA=1/2AB
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Answer 1

Answer:

hello

Step-by-step explanation:

proof :-

1. D is the mid point of AC

in ∆ ACB

M is mid point of AB

MD || BC

therefore, point D is mid point of AC (converse of mid point theorem)

AD =CD

2.MD perpendicular to BC

since, MD || BC

therefore,AC is a transversal.

therefore , angle BCD = angle MDA (corresponding angle)

angle BCD = 90°

angle MDA = 90°

therfore, MD perpendicular to AC.

3. CM= MA = 1/2 AB

join CM

In ∆ ADM & ∆ CDM

angle D = angle D ( Each 90°)

DM = DM (common side)

AD = CD ( D is mid point of AC)

therefore, ∆ADM congruent to ∆ CDM

( By SAS congruence rule)

CM = AM ( By CPCT)

Now,

AM = BM ( M is mid point of AB )

AM = BM = 1/2 AB

But,

AM = CM

So, AM = CM = 1/2 AB