Problem 2

In an acute-angled triangle

ABC

, the interior bisector of angle

A

intersects

BC

L

and intersects the circumcircle of

ABC

again at

N

. From point

L

, perpendiculars are drawn to

AB

and

AC

, with feet

K

and

M

respectively. Prove that the quadrilateral

AKNM

and the triangle

ABC

have equal areas.

Step 2 of 5: Classical bisector-chord identity via similar triangles

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In plain words

This is the standard 'power of the angle bisector' lemma: extending the bisector to the circle always turns the two triangles it cuts off into similar copies of each other.

AL \cdot AN = AB \cdot AC

Detailed analysis

Since $A,L,N$ are collinear, triangles $ABL$ and $ANC$ share the angle $\angle BAL=\angle NAC=A/2$ at $A$ ; also $\angle ABL=\angle ABC=\angle ANC$ because $\angle ABC$ and $\angle ANC$ subtend the same arc $AC$ of the circumcircle. Hence $\triangle ABL \sim \triangle ANC$ , giving $AB/AN = AL/AC$ , i.e. $AL\cdot AN = AB\cdot AC$ .