In the given figure the side BC of ∆ABC has been produced to point D. If the bisector of angle ABC and angle ACD meets at point E then prove that point E lie in the circumference o... In the given figure the side BC of ∆ABC has been produced to point D. If the bisector of angle ABC and angle ACD meets at point E then prove that point E lie in the circumference of a circle. Read more

Steps to Solve

1. Angle Bisector Theorem Application Since BE is the angle bisector of $\angle ABC$, we have $\angle ABE = \angle EBC$. Similarly, since CE is the angle bisector of $\angle ACD$, we have $\angle ACE = \angle ECD$.

2. Exterior Angle Property Using the exterior angle property of a triangle, we know that in $\triangle ABC$, $\angle ACD = \angle ABC + \angle BAC$.

3. Relating the Angles Since CE bisects $\angle ACD$, we can write $\angle ACE = \frac{1}{2} \angle ACD$. Substituting the expression for $\angle ACD$ from step 2, we get $\angle ACE = \frac{1}{2} (\angle ABC + \angle BAC)$.

4. Exterior Angle Property in Triangle EBC Now, consider $\triangle EBC$. The exterior angle property gives us $\angle ECD = \angle EBC + \angle BEC$. Since $\angle ECD = \angle ACE$ and $\angle EBC = \frac{1}{2} \angle ABC$, we can write $\angle ACE = \frac{1}{2} \angle ABC + \angle BEC$.

5. Equating and Simplifying Equating the two expressions for $\angle ACE$ from steps 3 and 4: $$ \frac{1}{2} (\angle ABC + \angle BAC) = \frac{1}{2} \angle ABC + \angle BEC $$ $$ \frac{1}{2} \angle ABC + \frac{1}{2} \angle BAC = \frac{1}{2} \angle ABC + \angle BEC $$ Subtracting $\frac{1}{2} \angle ABC$ from both sides, we are left with $\angle BEC = \frac{1}{2} \angle BAC$.

6. Angle at the Center Theorem Converse Let O be the circumcenter of $\triangle ABC$, and let the circle pass through A, B, and C. By the angle at the center theorem, $\angle BOC = 2 \angle BAC$. The angle subtended by the arc BC at the circumference is $\angle BAC$. In order for E to lie on the same circumference, $\angle BEC$ must equal $\angle BAC$ if E is on the major arc, or $180^\circ - \angle BAC$ if E is on the minor arc. Since $\angle BEC = \frac{1}{2} \angle BAC$, $\angle BAC = 2 \angle BEC$. This implies that $\angle BAC$ is twice $\angle BEC$. This corresponds to the angle subtended at the center by arc BC. Thus E lies on the circumference of the circle passing through points $A, B, C$