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In triangle $LMN$ and $LON$,

$LM=LO=5cm$

$MN=ON=5.5cm$

$LN=LN$ [Common]

$∴$ $△LMN≅△LON$ ...SSS criterion

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It is given, $AB=PQ$

$AC=PR$

$BC=QR$

Thus, $△ABC≅△PQR$ (SSS postulate)

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In triangle $PQR$ and $RSP$,

Given,

$SU=SU$

$ST=SV=5cm$

$UT=UV=3cm$

$∴$ $△SVU≅△STU$ .... SSS criterion

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$AD=BC$ (given);

$AB=AB$ (common);

$AC=BD$ (given).

We can use SSS condition to conclude that $△ABD≅△BAC$. From this we conclude that

$∠ADB=∠BCA$ and $∠DAB=∠CBA$.

In$ΔABC$ and $ΔPQR,AB=PQ,AC=PR$ and $BC=QR.$

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$AB=PQ$ [Given]

$AC=PR$ [Given]

$BC=QR$ [Given]

By Side - Side -Side criterion of congruency to each other

$ΔABC≅ΔPQR$

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$AB=DE$

$BC=EF$

$AC=DF$ or $CA=FD$

So option $C$ is correct.

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Therefore, D is the correct answer.

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In triangle $AOB$ and $DOE$,

Given,

$AO=DO=2cm$

$DE=AE=2cm$

$EO=BO=1.5cm$

$∴$ $△AOB≅△DOE$ .. SSS criterion

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AB=DC , AC=DB and BC=BC

then from $SSS$ congruence ,$ΔABC=ΔDCB$

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