maths

To prove: $BE=CF$

In $△$s BEC and BFC

$∠BEC=∠BFC=90$ (Given)

BC = BC (common)

$∠ABC=∠ACB$ (Given)

Thus, $△$s BEC $≅$ CFB (AAS Congruency)

Hence, $BE=CF$

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Prove that :

$AM=AN$

$ΔAMC≅ΔANB$

$BN=CM$

$ΔBMC≅ΔCNB$

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(i) $△ABD≅△ACD$

(ii) $△ABP≅△ACP$

(iii) $AP$ bisects $∠A$ as well as $△D$.

(iv) $AP$ is the perpendicular bisector of $BC$.

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Hence, The triangles ADX and BAY are congruent

If the above statement is true then mention answer as 1, else mention 0 if false

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