Question

0. In Fig. 6.109, the side BC of a AABC is produced. The bisector of ZBAC intersects the side BC at 5;
[CBSE 2010, '12]
Prove that ZABC + ZACD = 2 angle AEC

AP​

Answer 1

$\Huge{\red{\underline{\textsf{Answer}}}}$

$\large{\green{\underline{\tt{Given}}}}$

The side BC of ∆ ABC is produced to D, the bisector of $\angle BAC$ intersects the side BC at E.

$\large{\blue{\underline{\tt{To\:prove}}}}$

$\angle ABC$ + $\angle ACD$ = 2$\angle AEC$

$\large{\purple{\underline{\tt{Proof}}}}$

We know that $\angle ACD$ = $\angle ABC$ + $\angle BAC$....(1)

{ext. angle theorem}

AE is the bisector of $\angle BAC$, thus

i.e., $\angle BAC$ = 2$\angle BAE$

Thus equation (1) reduce to

$\leadsto$$\angle ACD$ = $\angle ABC$ + 2$\angle BAE$

Adding $\angle ABC$ on both sides,

Now $\angle ABC$ + $\angle ACD$=$\angle ABC$ + $\angle ABC$ + 2$\angle BAE$

$\leadsto$2($\angle ABC$ + $\angle BAE$)

$\leadsto$2($\angle AEC$) {ext. angle of ∆ ABE}

Hence proved.

❤️ $\mathfrak{Be \: Brainly}$❤️