In the given figure,the side BC of ∆ABC has been produced to a point

1.	In the given figure,the side BC of ∆ABC has been produced to a point D.If the bisectors of ∠ABC and ∠ACD meet a point E then prove that ∠BEC =½ ∠BAC.
Answer» ━━━━━━━━━━━━━━━━━━━━━━━━━ $\huge{\underline{\underline{\sf{\orange{Question:-}}}}}$ In the given figure,the side BC of ∆ABC has been produced to a POINT D.If the bisectors of ∠ABC and ∠ACD meet a point E then PROVE that ∠BEC =½ ∠BAC. ━━━━━━━━━━━━━━━━━━━━━━━━ $\huge{\underline{\underline{\sf{\orange{Solution:-}}}}}$ Side BC of ∆ABC has been produced to D. ∴ ∠ACD = ∠BAC + ∠ABC $⇒ \frac{1}{2} \angle \: ACD = \frac{1}{2} \angle \: bAC \: + \frac{1}{2} \angle \: ABC \\$ $⇒ \angle \: ECD = \frac{1}{2} =\angle \: BAC + \frac{1}{2} \angle \: ABC. \: \: \: \: \: \: \: \: ....(i)\\$ Again,side BC of ∆EBC has been produced to D. $\therefore \: \angle \: ECD = \angle \: CBE + \angle \: BEC$ $⇒ \angle \: ECD = \frac{1}{2} \angle \: ABC + \angle \: BEC. \: \: \: \: \: \: \: ...(ii) \\$ From (i) and (ii), we GET $\frac{1}{2} \angle \: ABC + \angle \: bEC = \frac{1}{2} \angle \: BAC + \frac{1}{2} \angle \: ABC \\ \\ \: \: \: \: \: \: \: \: (each \: equal \: to \: \angle \: ECD).$ $\therefore \: \angle \: BEC = \frac{1}{2} \angle \: BAC.$ ━━━━━━━━━━━━━━━━━━━━━━━━━

In the given figure,the side BC of ∆ABC has been produced to a point D.If the bisectors of ∠ABC and ∠ACD meet a point E then prove that ∠BEC =½ ∠BAC.

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$\huge{\underline{\underline{\sf{\orange{Question:-}}}}}$

In the given figure,the side BC of ∆ABC has been produced to a POINT D.If the bisectors of ∠ABC and ∠ACD meet a point E then PROVE that ∠BEC =½ ∠BAC.

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$\huge{\underline{\underline{\sf{\orange{Solution:-}}}}}$

Side BC of ∆ABC has been produced to D.

∴ ∠ACD = ∠BAC + ∠ABC

$⇒ \frac{1}{2} \angle \: ACD = \frac{1}{2} \angle \: bAC \: + \frac{1}{2} \angle \: ABC \\$

$⇒ \angle \: ECD = \frac{1}{2} =\angle \: BAC + \frac{1}{2} \angle \: ABC. \: \: \: \: \: \: \: \: ....(i)\\$

Again,side BC of ∆EBC has been produced to D.

$\therefore \: \angle \: ECD = \angle \: CBE + \angle \: BEC$

$⇒ \angle \: ECD = \frac{1}{2} \angle \: ABC + \angle \: BEC. \: \: \: \: \: \: \: ...(ii) \\$

From (i) and (ii), we GET

$\frac{1}{2} \angle \: ABC + \angle \: bEC = \frac{1}{2} \angle \: BAC + \frac{1}{2} \angle \: ABC \\ \\ \: \: \: \: \: \: \: \: (each \: equal \: to \: \angle \: ECD).$

$\therefore \: \angle \: BEC = \frac{1}{2} \angle \: BAC.$

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In the given figure,the side BC of ∆ABC has been produced to a point D.If the bisectors of ∠ABC and ∠ACD meet a point E then prove that ∠BEC =½ ∠BAC.

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In the given figure,the side BC of ∆ABC has been produced to a point D.If the bisectors of ∠ABC and ∠ACD meet a point E then prove that ∠BEC =½ ∠BAC.​

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In the given figure,the side BC of ∆ABC has been produced to a point D.If the bisectors of ∠ABC and ∠ACD meet a point E then prove that ∠BEC =½ ∠BAC.