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  • Two sides AB and BC, and the median AD of ABC are correspondingly equal to the two sides PQ and QR, and the median PM of

Two sides AB and BC, and the median AD of \DeltaABC are correspondingly equal to the two sides PQ and QR, and the median PM of \DeltaPQR. Then,

Option: 1

\angle A B C=\angle P Q R


Option: 2

BD=QM


Option: 3

\triangle A B C \cong \Delta P Q R


Option: 4

All of the above


Answers (1)

best_answer

\\\text{In } \triangle A B D\text{ and }\triangle P Q M,\text{ we have} \\\\A B=P Q \quad(\text { given }) \\ \\B D=Q M \quad\left[\because B C=Q R \Rightarrow \frac{1}{2} B C=\frac{1}{2} Q R \Rightarrow B D=Q M\right] \\\\\text { med. } A D=\text { med. } P M \quad(\text { given })

\\\therefore \quad \triangle A B D \cong \triangle P Q R \quad \text { (SSS-criteria) } \\\\ \therefore \quad \angle A B D=\angle P Q M \quad( \text { c.p.c.t.) } \\\\ \Rightarrow \quad \angle A B C=\angle P Q R\\ \\\text{Now, in } \triangle A B C\text{ and } \triangle P Q R,\text{ we have} \\\\ A B=P Q \quad(\text { given })

\\ \angle A B C=\angle P Q R \quad \text { (proved above) } \\ \\B C=Q R \quad \text { (given). } \\ \\\therefore \quad \triangle A B C \cong \Delta P Q R \quad \text { (SAS-criteria). }

 

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HARSH KANKARIA

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