Question

$∠DAB=∠DEC$

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Given ; A $ΔABC$ in which D is the mid - point of BC

AD is produced to E so that DE = AD

We need to prove that

Also , $∠DAB=∠DEC$ [The corresponding parts of congruent triangles are equal]

AD is produced to E so that DE = AD

We need to prove that

$ΔABD≅ΔECD$

AB = EC

AB \\ EC

In $ΔABD$ and $ΔECD$

BD = DE [D is the midpoint of BC]

$∠ADB=∠CDE$ [Vertically opposite angles]

AD = DE [Given]

By Side - Angle - Side criterion of congruence we have

$ΔABD≅ΔECD$

AB \\ EC

In $ΔABD$ and $ΔECD$

BD = DE [D is the midpoint of BC]

$∠ADB=∠CDE$ [Vertically opposite angles]

AD = DE [Given]

By Side - Angle - Side criterion of congruence we have

$ΔABD≅ΔECD$

Also , $∠DAB=∠DEC$ [The corresponding parts of congruent triangles are equal]

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