Question #89363
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
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Expert's answer
2019-05-08T14:21:43-0400

Consider ΔABC.\Delta ABC.



Given that


BDDC=BFAF{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 FDACFD \Vert AC


FDA=CAD\angle FDA=\angle CAD

Consider a convex quadrilateral AFKE.AFKE.


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

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


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

BFK+KDB=180\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.


BFK+KDB=180=FBD+FKD\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


FBK=FDK\angle FBK=\angle FDK

ABE=FDA\angle ABE=\angle FDA

Therefore,


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

Therrefore,


ABE=CAD\angle ABE=\angle CAD

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Comments

Assignment Expert
09.05.19, 17:02

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Aravind
09.05.19, 16:07

Thanks

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