Answer to Question #89363 in Trigonometry for Aravind

Question

Let ABC be a triangle,let D,E,F be points on BC,CA and AB such the AD,BE and CF are concurrent at the point K. Suppose BD/DC=BF/AF and angle ADB= angle AFC. Prove that angle ABE=angle CAD
Please explain in simple way

Answer 1

Consider $\Delta ABC.$

Given that

$${BD \over DC}={BF \over AF}$$

Converse of Triangle Proportionality Theorem

If a line divides two sides of a triangle proportionally then it is parallel to the third side.

Hence $FD \Vert AC$

$$\angle FDA=\angle CAD$$

Consider a convex quadrilateral $AFKE.$

$$\angle BFC+\angle AFC=180^{\circ}$$

Given that $\angle ADB=\angle AFC.$ Then

$$\angle BFC+\angle ADB=180^{\circ}$$

$$\angle BFK+\angle KDB=180^{\circ}$$

The measure of the interior angles of a convex quadrilateral is the same as the sum of the measures of the interior angles of two triangles, or 360 degrees.

$$\angle BFK+\angle KDB=180^{\circ}=\angle FBD+\angle FKD$$

If two opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.  

From Inscribed Angle Theorem we have

$$\angle FBK=\angle FDK$$

$$\angle ABE=\angle FDA$$

Therefore,

$$\angle ABE=\angle FDA=\angle CAD$$

Therrefore,

$$\angle ABE=\angle CAD$$