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Question 8 (iii)
In right 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 (see the given figure). Show that:
ΔDBCΔACB

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Solution

In ΔAMC andΔBMDAM=BM (M is midpoint of AB)AMC=BMD(vertically opposite angles)CM=DM (given)ΔAMCΔBMD (by SAS congruence rule) AC=BD (by CPCT)And ACM=BDM(by CPCT) We have ACM=BDMBut ACM and BDM are alternate interior anglesSince alternate angles are equal.Hence, we can say that DB||ACDBC+ACB=180(cointerior angles) DBC+90=180(ACB=90)DBC=18090DBC=90DB=AC (By CPCT)(i)In ΔDBC and ΔACBDB=AC (From (i))BC=BC(Common)DBC=ACB=90ΔDBCΔACB by SAS congruence

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