ABC is a triangle. BB$_1$ and CC$_1$ are angle bisectors of B and C respectively. E,F are feet of perpendiculars from A on BB$_1$ and CC$_1$ respectively. Suppose D is point at which incircle of ABC touches AB, Prove that AD = EF.
Use coordinate geometry only.
On
Synthetic solution (out of habit, angles are measured in degrees):
Lemma: ADFIE are concyclic points.
Proof: $AE \perp EI$ and $AF \perp FI$, so $AEFI$ is a cyclic quadrilateral. $AD \perp DI$ and $AF \perp FI$, so $ADFI$ is a cyclic quadrilateral.
Lemma: $AE \parallel DF$.
Proof: First notice $\angle BAE = 90 - \frac{1}2 \angle ABC$.
Then $\angle FDA = \angle AIC$ by cyclicity of the quadrilateral $ADFI$, and $\angle AIC = 180 - \frac12 \angle BAC - \frac12 \angle CAB.$ Therefore the complementary angle $\angle BDF$ has measure $\frac{1}2 \left( \angle BAC + \angle CAB \right) = 90 - \frac12 \angle ABC = \angle BAE$. It follows $DF \parallel AE$.
Then, $AE$ and $DF$ are parallel lines cutting the circle formed by A,D,F,I, and E. Therefore the segments AD and FE have equal length.
NOTE : This is not a complete answer. This is the answer for the case that $ABC$ is an isosceles triangle. You'll see how ugly it becomes. If you understand how it goes, you can generalize this answer.
Without loss of generality, we may suppose that $A(0,a),B(-1,0),C(1,0)$ where $a\gt0$. (Do you see why?)
Letting $\tan\theta=1/a$, which is the slope of the line $AB$, the slope of the line $BB_1$ is $$\tan\frac{\theta}{2}=\sqrt{\frac{\sqrt{a^2+1}-1}{\sqrt{a^2+1}+1}}=a^\prime.$$ In the same way, you know that the slope of the line $CC_1$ is $$-\sqrt{\frac{\sqrt{a^2+1}+1}{\sqrt{a^2+1}-1}}=a^{\prime\prime}.$$ Hence, the line $BB_1, CC_1$ can be represented as $$BB_1 : y=a^\prime x+a^\prime, \ CC_1 : y=a^{\prime\prime}x-a^{\prime\prime}.$$ By the way, the lines $AF,AE$ can be represented as $$AF : y=-\frac{1}{a^{\prime\prime}}x+a, \ AE : y=-\frac{1}{a^{\prime}}x+a.$$ Hence, we can get the coordinate of $F,E$ as $$F \left(\frac{a^{\prime\prime}+a}{a^{\prime\prime}+(1/{a}^{\prime\prime})},\frac{aa^{\prime\prime}-1}{a^{\prime\prime}+(1/{a}^{\prime\prime})}\right), E \left(\frac{a-a^{\prime}}{a^{\prime}+(1/{a}^{\prime})},\frac{aa^{\prime}+1}{a^{\prime}+(1/{a}^{\prime})}\right).$$
By the way, since the incenter $I$ of the triangle is the intersection point of the lines $BB_1, CC_1$, we get $$I \left(\frac{a^{\prime\prime}+a^{\prime}}{a^{\prime\prime}-a^{\prime}},\frac{2a^{\prime}a^{\prime\prime}}{a^{\prime\prime}-a^{\prime}}\right).$$ NOTE : $a^{\prime}\not=a^{\prime\prime}$ because $a^{\prime}\gt0, a^{\prime\prime}\lt0.$
Hence, we know the point $D$ is the intersection point of $AB : y=ax+a$ and the line which is perpendicular to $AB$ and passes through $I$. Then, the latter line can be represented as $$y-\frac{2a^{\prime}a^{\prime\prime}}{a^{\prime\prime}-a^{\prime}}=-\frac 1ax\left(x-\frac{a^{\prime\prime}+a^{\prime}}{a^{\prime\prime}-a^{\prime}}\right).$$ Hence, we get $$D \left(\frac{a(2a^{\prime}a^{\prime\prime}+aa^{\prime}-aa^{\prime\prime})}{(a^2+1)(a^{\prime\prime}-a^{\prime})},\frac{a(2aa^{\prime}a^{\prime\prime}-a^{\prime}+a^{\prime\prime})}{(a^2+1)(a^{\prime\prime}-a^{\prime})}\right).$$
Now, in order to prove $AD^2=EF^2$, all we need is to check if $$\left(A_x-D_x\right)^2+\left(A_y-D_y\right)^2=\left(E_x-F_x\right)^2+\left(E_y-F_y\right)^2$$ holds for any $a\gt 0$. This is just a calculation, so I left it to you.