Question

Let $ABC$ be a triangle. Let $D,E,F$ be the points respectively on the segments $BC,CA,AB$ such that $AD,BE,CF$ concur at the point $K.$ Suppose $\frac{BD}{DC}=\frac{BF}{FA}$ and $\angle ADB=\angle AFC$. Is $\angle ABE=\angle CAD.?$

• A. True
• B. False

Answer 1

The correct option is A True

As It is given ,

$\frac{BD}{DC}=\frac{BF}{FA}$ and $\angle ADB=\angle AFC$

The line DF and CA are parallel

We have $\angle BDK=\angle ADB=180-\angle BFK$

So that BDFK is cyclic quadrilateral

$\therefore \angle FBK=\angle BDK$

$\angle ABE=\angle FBK=\angle FDK=\angle FDA=\angle DAC$

finaly $\angle ABE=\angle CAD$