Let a circle with diameter $AB$ have points $C$ and $D$ on same side so that $BD$ bisects $\angle CBA$. Chords $AC$ and $BD$ intersect at $E$.

Question

Let a circle with diameter $AB$ have points $C$ and $D$ on same side so that $BD$ bisects $\angle CBA$. Chords $AC$ and $BD$ intersect at $E$. Given that $AE = 169$ cm and $EC = 119$ cm, find $ED$.

What I Tried: Here is a picture :-

We have $AE = 169, EC = 119$.
Let $AB = r$ , $DE = x$ , $AD = y$ , $BC = z$ , $BE = a$.
Now, applying Pythagoras Theorem gives all these equations one by one :- $$(i) x^2 + y^2 = 169^2$$ $$(ii) z^2 + 119^2 = a^2$$ $$(iii) (x+a)^2 + y^2 = R^2$$ $$(iv) 288^2 + z^2 = R^2$$ We also have $\Delta CEB \sim \Delta DAB \sim \Delta DAE$ . They give information as :- $$(i) \frac{CE}{DA} = \frac{CB}{DB} = \frac{EB}{AB}$$ $$(ii) \frac{DE}{EC} = \frac{DA}{CB} = \frac{AE}{EB}$$

The Problem is that all these don't really make me find the value of $ED$, or I am unable to find it by using them appropriately in some way.

Can anyone help me? Thank You.

Answer 1

let $Q$ be the foot of the perpendicular from $E$ to $AB$ then $EC=EQ$ , $AQ^2 = AE^2-EQ^2 $ and $$\frac{AQ} {EQ} =\frac{AC} {CB}$$ You can calculate $EB$ by pythagoras from $\triangle EBC$ , and finally $$AE \cdot EC= EB \cdot DE$$

Answer 2

By angle bisector rule $$\frac{CB}{AB}=\frac{119}{169}$$ also by Pythagoras $$AB^2+CB^2={(169+119)}^2$$ $$AB=?,CB=?$$ Now in $\Delta CEB $ $$\tan x=\frac{119}{CB}=?$$ From here you can find $\sin x$ but $$\sin x=\frac{DE}{169}$$

We are done!

Answer 3

Let $AB = 169y$, $BC=119y$.

$$AC^2=AB^2-BC^2\Rightarrow y^2=\ldots$$

$$BE^2=CE^2+BC^2\Rightarrow BE^2=\ldots$$

$$DE\cdot BE = AE\cdot CE \Rightarrow DE=\ldots$$

Answer 4

Note $ DE \cdot EB = AE \cdot EC =169\cdot 119 $, or $$DE \cdot (DB -DE) =169\cdot 119$$ Substitute $$DB = AB \cos x= AB \cdot\frac{\sin 2x}{2\sin x}=AB \cdot\frac{\frac{AC}{AB}}{2\frac{DE}{AE}}= \frac{144\cdot169}{DE}$$ to get $$DE^2= 169\cdot(144-119)=169\cdot 25=(13\cdot5)^2$$ Thus, $DE=65$.