Answer to Question #128514 in Geometry for Precious Adeleye

Question

The bisector of the angle A at the base of the trapezoid ABCD is perpendicular to it's diagonal and crosses the side CD,at the point E and CE : ED=2:3,Find the ratio of BC:CK where k is the intersection point of the bisector with line BC.

Answer 1

Consider a trapezoid ABCD shown in the diagram above.

Let AE be the bisector of $\angle A$ where $\angle DAE = x$ so $\angle BAK = x$ as per the angular bisector property.

As per the question AE is perpendicular to the diagonal BD

$\therefore$ $\angle ADB = \angle ABD = 90\degree - x$

$\angle KCE = x$ as AE is intersecting two parallel lines AB and DE so incident angles are equal.

Let $\angle CKE = a$ so $\angle BKA = a$ as vertically opposite angles are equal.

$\therefore$ we can say that $\triangle ABK \cong \triangle CEK$ by AAA congruent property.

$\therefore$ CE/AB = CK/BK

Now we can write BK as (BC-CK).

Also AB= AD as $\triangle ADB$ is an isosceles triangle.

Also $\triangle ADE$ is an isosceles triangle because two base angles are equal to $x$

so AD = ED which means AB = ED

Now coming back to below equation and putting the values we get.

CE/AB = CK/BK

CE/ED = CK/(BC-CK)

ED/CE = (BC-CK)/CK

ED/CE = (BC/CK) - 1

From the question ED/CE = 3/2

$\therefore$ BC/CK = (3/2) + 1

= 5/2