Q: 8     In Fig. \small 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment \small AX\perp DE  meets BC at Y. Show that:

    (i)    \small \Delta MBC\cong \Delta ABD

          

Answers (1)


We have, a \DeltaABC such that BCED, ACFG and ABMN are squares on its side BC, CA and AB respec.  Line segment \small AX\perp DE  meets BC at Y

(i) \angle CBD = \angle MBA   [each 90] 
Adding \angle ABC on both sides, we get
\angle ABC + \angle CBD = \angle MBA + \angle ABC
\Rightarrow \angle ABD = \angle MBC
In \DeltaABD and \DeltaMBC, we have
AB = MB
BD = BC
\angle ABD = \angle MBC
Therefore, By SAS congruency 
\DeltaABD \cong \DeltaMBC

Hence proved.

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