Answer:
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Step-by-step explanation:
In triangles ADC and CBA, we have
AB=CD ......... (given);
AD=BC ......... (given);
AC=AC ......... (common side).
By S.S.S congruency condition,
△ADC≅△CBA [henceproved]
In △PQR and △SRQ,
PQ=SR(Given)
QR=QR(Common)
PR=SQ(Given)
By SSS property,
△PQR≅△SRQ
Hence proved.
△ABD and △ACD
AB=AC (given )
Then ∠ABD=∠ACD ( because AB=AC )
and ∠ADB=∠ADC=90( because AD⊥BC )
∴△ABD=△ACD
∠BAD=∠CAD
It is defective to use ∠ABD=∠ACD for proving this result
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Answers & Comments
Verified answer
Answer:
yeah I'm here sis!!❣️
Step-by-step explanation:
ANS1:
Solution:
In triangles ADC and CBA, we have
AB=CD ......... (given);
AD=BC ......... (given);
AC=AC ......... (common side).
By S.S.S congruency condition,
△ADC≅△CBA [henceproved]
ANS2:
Solution:
In △PQR and △SRQ,
PQ=SR(Given)
QR=QR(Common)
PR=SQ(Given)
By SSS property,
△PQR≅△SRQ
Hence proved.
ANS3:
Solution:
△ABD and △ACD
AB=AC (given )
Then ∠ABD=∠ACD ( because AB=AC )
and ∠ADB=∠ADC=90( because AD⊥BC )
∴△ABD=△ACD
∠BAD=∠CAD
It is defective to use ∠ABD=∠ACD for proving this result