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  • In Fig. 6.37, if D ABE equivalent Delta ACD, show that D ADE ~ D ABC.

Q6  In Fig. 6.37, if \Delta ABE \equiv \Delta ACD, show that  \DeltaADE ~ \Delta ABC.    

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Given : \triangle ABE \cong \triangle ACD

To prove ADE ~ \Delta ABC.

Since \triangle ABE \cong \triangle ACD

 AB=AC          ( By CPCT)

  AD=AE          (By CPCT)

In \DeltaADE  and  \Delta ABC,

 \angle A=\angle A    ( Common)

and    

  \frac{AD}{AB}=\frac{AE}{AC}  (  AB=AC   and   AD=AE )

Therefore, \DeltaADE ~ \Delta ABC.   ( By SAS criteria)

 

 

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