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Question

In right angled triangle ABC, right angled at C,M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM=CM. Point D is joined to point B. Show that:
(i) AMCBMD
(ii) DBC is a right angle.
(iii) DBCACB
(iv) CM=12AB
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Solution

(i) In AMC and BMD, we have

AM=BM Since M is the mid- point of AB

AMC=BMD Vertically opp. angles
and CM=MD Given

By SAS criterion of congruence, we have
AMCBMD

(ii) Now, AMCBMD

BD=CA and BDM=ACM ... (1) Since corresponding parts of congruent triangles are equal

Thus, transversal CD cuts CA and BD at C and D respectively such that the alternate angles BDM,ACM are equal.

,BDCA.

CBD+BCA=180 Since sum of consecutive interior angles are supplementary

CBD+90=180 Since BCA=90

CBD=18090

DBC=90

(iii) Now, in DBC and ACB, we have

BD=CA From (1)

DBC=ACB Since Each =90o

BC=BC Common

by SAS criterion of congruence, we have
DBCACB

(iv) CD=AB [Since corresponding parts of congruent triangles are equal]

M is the midpoint of CD.

CM=12CD

CM=12AB

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