Let $M$ be the midpoint of $AL$, let $\angle{ABL}=\angle{LBC}=x,\angle{MEL}=y$. Since $\frac{ML}{ME}=\frac{MA}{ME}$, we get $\frac{\sin (y)}{\sin (2x-y)}=\frac{\sin (2x)}{\sin (4x)}=\frac{1}{2\cos (2x)}$. So, $2\cos (2x)\sin (y)=\sin (2x-y)\implies 3\cos (2x)\sin (y)=\sin (2x)\cos (y)$. We also have $\frac{CD}{CL}=\frac{AL}{2CL}=\frac{AB}{2BC}=\frac{\sin (2x)}{2\sin (4x)}=\frac{1}{4\cos (2x)}$. Note that $4\sin (y)\cos (2x)=\sin (2x)\cos (y)+\cos (2x)\sin (y)=\sin (2x+y)$. So $\frac{1}{4\cos (2x)}=\frac{\sin (y)}{\sin (180^{\circ}-2x-y)}=\frac{\sin (\angle{CLD})}{\sin (180^{\circ}-2x-\angle{CLD})}$. Hence, $\angle{CLD}=y$ and $\angle{CDL}=180^{\circ}-2x-y=\angle{LEB}$. So, $BELD$ lie on a circle, and so $LE=LD$.

Project $L$ onto $AB$ and $BC$ at $Q$ and $P$. Obviously , $LP = LQ$ , $\angle {LQE} = \angle {LPD} = \frac{\pi}{2}$. After using The Sine Law and The Bisector's Theorem , we get that showing $QE = PD$ is equivalent to showing : $ 4 \cdot sin^4 (a) + cos (2a) = cos^2 (2a) + 2 \cdot sin^2(a) $ , for some $a$ with $ 0 < a < \frac{\pi}{2} $ ( where $\angle{BAC} = 2a$ ) , which is easy. Now , $QE = PD$ , so $ \triangle{LQE} = \triangle{LPD} $ , which means that $LE = LD$. Done.

Let $F$ point on $AB$ such that $LF \parallel BC$ Then $\frac{LF}{BC}=\frac{LA}{AC}$ and $\frac{LC}{BC}=\frac{LA}{AB } \to LF=LC$ Let $M,E$ are midpoints of $AL,AF$. Then $EF=AM=CD$ and $\angle LCD = \angle AFL$ so $\triangle LCD= \triangle EFL$ so $LE=LD$

AngleChasingXD wrote: $BL $ is the bisector of an isosceles triangle $ABC $. A point $D $ is chosen on the Base $BC $ and a point $E $ is chosen on the lateral side $AB $ so that $AE=\frac {1}{2}AL=CD $. Prove that $LE=LD $. Tuymaada 2017 Q5 Juniors My solution: We know that $AL=\dfrac{b^2}{a+b}, CL=\dfrac{ab}{a+b}$. So $AE=CD=\dfrac{b^2}{2(a+b)}$. Using law of cosine in $\triangle CDL $ : $ LD^2=\dfrac{(ab)^2}{(a+b)^2}+\dfrac{b^4}{4(a+b)^2}-2\cdot \dfrac{b^2}{2(a+b)} \cdot \dfrac{ab}{a+b} \cdot \cos{C}=\dfrac{b^2(2a^2+b^2)}{4(a+b)^2}$ By law of cosine in $\triangle AEL : LE^2=\dfrac{b^4}{(a+b)^2} +\dfrac{ b^4}{4(a+b)^2} -2 \cdot \dfrac{b^2}{a+b} \cdot \dfrac{b^2}{2(a+b)} \cdot \cos{A} = \dfrac{5b^4-b^4+2a^2b^2}{4(a+b)^2}=\dfrac{b^2(2a^2+b^2)}{4(a+b)^2} $ So $LE=LD$

Why? Why do you guys prefer an ugly method. Draw a circle passing through points \(A , L , D\) to intersect \(BC\) at \(K.\) Use chicken feet theorem. Everything becomes clear I will complete my post

RC. wrote: Why? Why do you guys prefer an ugly method. Draw a circle passing through points \(A , L , D\) to intersect \(BC\) at \(K.\) Use chicken feet theorem. Everything becomes clear I will complete my post Lo and behold, the hero never returned.

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thats enough to prove that $BELD$ is cyclic. let $N$ be the midpoint of $AC$ . thats enough to prove that $BNLD $is cyclic which this is obvious(by little computing)

Let S be on AB such that SL || BC. we will prove triangles LES and LDC are congruent. CL/LA = BC/BA so CL/BC = LA/BA = AS/AB = SL/BC so SL = CL. ES = DC , LS = LC and ∠LCD = ∠LSE so LES and LDC are congruent. now we know LE = LD. we're Done.