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Question

In a right ABC, right angled at C, M is the mid point of the hypotenuse of AB. C is joined to M produced to a point D such that DM=CM. Point D is joined to point B. Then
244457.png
  1. AMCBMD
  2. DBCACB
  3. DBC is a right angle
  4. CM=12AB

A
DBC is a right angle
B
CM=12AB
C
AMCBMD
D
DBCACB
Solution
Verified by Toppr

Option A:
In AMC and BMD
AM=BM ....Since M is the midpoint of the hypotenuse AB
CM=DM .... Given
AMC=BMD ..... Vertically opposite angles
AMCBMD .... SAS Rule

Option B:
ACM=BDM .... cpct
But these are the alternate angles and they are equal.
AC||BD
Now, AC||BD and a transversal BC intersects, then
DBC+ACB=180o
The sum of consecutive interior angles on the same sides of the tranversal is 180o.
DBC+90o=180o
DBC=90o
DBC is a right angle.

Option C:
In DBC and ACB
DBC=ACB=90o
BC=CB .... Common side
AMCBMD ... Proved
AC=BD .... cpct
DBCACB ..... SAS Rule

Option D:
DBCACB .... Proved
DC=AB .... cpct
2CM=AB .... DM=CM=12DC
CM=12AB

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