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:

(iv)  \small \Delta FCB\cong \Delta ACE

        

Answers (1)


\angle FCA = \angle BCE  [both 90]
By adding \angleABC  on both sides we get
\angleABC + \angleFCA = \angleABC + \angleBCE
\angleFCB = \angleACE

In \DeltaFCB and \DeltaACE
FC = AC [sides of square]
BC = AC [sides of square]
\angleFCB = \angleACE
\Rightarrow 
\DeltaFCB  \cong  \DeltaACE

Hence proved

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